Arithmetic of finance

1. Revision

Finding the length of an investment

At some time in the past, Mike deposited R2 600 into a savings account at The Bank of Money Money. The account earns interest at a rate of 6,82% compounded yearly. How long ago did Mike open the account if the balance is now R3 204,10? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
What kind of interest was earned? Which formula do we need for this calculation?
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

A=P(1+i)nWhere: A=3 204,1P=2 600i=6,82%=0,0682n=?

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for n.

3 204,1=2 600(1+0,0682)n3 204,1=2 600(1,0682)n3 204,12 600=(1,0682)n

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that n is the subject of the formula:

n=log1,0682(3 204,12 600)n=3,16666...

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Mike left the money in the account for about 3,17 years. However, we need to give the answer in terms of years and months.

3,16666... years = 3 years and 0,16666... of a year (which will be in months).

We know that there are 12 months in a year. To convert 0,16666... of a year into months, we do the following calculation:

0,16666... of a year×(12 monthsyear)=2 month(s)

Therefore, Mike deposited the money into the account 3 years and 2 month(s) ago.


Submit your answer as: and

Timeline question: adding or withdrawing money

Port Elizabeth First Bank offers a savings account which earns 6,2% interest p.a. compounded once each month. Babangida decides to open an account, and deposits R4 000. His account accrues interest for 8 years until he puts R3 000 more into the account. After that, Babangida leaves the money in the account until 10 years after his original deposit. How much money will be in the account in the end?

Answer: Total amount = R
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.


STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of 10 years. This period is broken into two sections: 8 years after Babangida opens his account, and then the last 2 years after he puts R3 000 more into his account. We must calculate the value in the account separately for these two sections of time.

For the first 8 years we have:

P=R4 000i=6,2%=0,062n=8A= ?

The question tells us that the interest rate is "6,2% interest p.a. compounded once each month" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by 12 and divide the interest rate by 12 because the interest compounds once each month.

A=P(1+i)nA=4 000(1+0,06212)(8×12)=4 000(1,005...)96=4 000(1,640...)=R6 560,18The amount after 8 years.

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Babangida puts R3 000 more into his account: did Babangida put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

P=6 560,18+3 000=R9 560,18i=6,2%=0,062n=2A= ?

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last 2 years. (As above, we must adjust the exponent and the interest rate.)

A=9 560,18(1+0,06212)(2×12)=9 560,18(1,005...)24=9 560,18(1,131...)=R10 818,82The amount after 10 years.

In the end, Babangida will have R10 818,82 in his account at the bank.


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Finding the length of an investment

Mpumelelo puts R350 into a bank account at the First Bank of Springbok. Mpumelelo's account earns interest at a rate of 7,31% p.a. compounded four times per year. After how many years will the bank account have a balance of R647,86?

INSTRUCTION: Give your answer to one decimal place.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable four times per year, as this affects i and n in the compound interest formula.


STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

A=P(1+i)nWhere: A=647,86P=350i=7,31%=0,0731n=?

STEP: Adjust the values of i and n based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded four times per year.

We must adjust the values of i and n according to the compounding period. For this question, we must make the following adjustments:

i0,07314 and n(n×4)

In this case, n represents the number of years and (n×4) represents the number of times the bank pays interest into the account.

647,86=350(1+0,07314)(n×4)

Simplify the equation:

647,86=350(1,01827...)4n647,86350=(1,01827...)4n

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that 4n becomes the subject of the formula:

4n=log1,018275(647,86350)4n=34

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for n:

4n=34n=344n=8,5

Therefore, the money has been in the Mpumelelo's account for 8,5 years.

NOTE: If you got an answer close to this (for example, 8,62 years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is 8,5.


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Timelines

Chijindum wants to buy a used car. The cost of the used car is R67 750. In 2003 Chijindum opened an account at Sutherland Stars Bank with R18 000. Then in 2007 he withdrew R3 250 from the account. In 2013 Chijindum made another change: he put R3 000 into the account. If the account pays 7,54% p.a. compounded every six months, how much money does he have now, and is it enough to buy the used car?

Answer:
  1. The total amount in the account now is R .
  2. Does he have enough money?
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2003 to 2007 is a period of 3 years.


STEP: Find the amount of money there was in 2007 (starting in 2003)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2003 to 2007:

P=R18 000i=7,54%=0,0754n=20072003=4 yearsA= ?

Solve the first part of the problem:

A=18 000(1+0,07542)(4×2)A=18 000(1,0377)8=18 000(1,34454...)=R24 201,76The amount in 2007.

STEP: Adjust the amount of money to reflect Chijindum's withdrawal from the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2007 to 2013. You must adjust the amount of money in the account if Chijindum put more money in or took money out.

P=24 201,763250=R20 951,76i=7,54%=0,0754n=20132007=6 yearsA= ?

Now solve this part of the problem:

A=20 951,76(1+0,07542)(6×2)=20 951,76(1,0377)12=20 951,76(1,55905...)=R32 664,97The amount in 2013.

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2013 to 2018: did Chijindum put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

P=32 664,97+3000=R35 664,97i=7,54%=0,0754n=20182013=5 yearsA= ?

Now solve the third part of the problem, from 2013 to 2018:

A=35 664,97(1+0,07542)(5×2)=35 664,97(1,0377)10=35 664,97(1,44783...)=R51 636,88The amount this year.

STEP: State the final answer (including whether or not Chijindum has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Chijindum to buy the used car.

The answers are:

  1. This year Chijindum has R51 636,88 in his account, and so
  2. no, he can not afford the used car.

Submit your answer as: and

Timeline questions: following the changes

At the beginning of 2006 a young woman starts a savings account at Cape Town Bank. She invests R7 150 into the account. The interest rate is 8,96 % p.a. compounded daily. In 2009 the bank changes the interest rate to 8,33 % p.a. compounded daily. Then in 2011, the interest rate changes again to 8,33 % p.a. compounded weekly.

  1. How much money will the woman have in her account in 2015, 9 years after her original deposit?
  2. How much interest will she earn from the bank during this time?
INSTRUCTION: Assume that every year has 365 days (ignore leap years).
Answer:
  1. She will have R in the account.
  2. She will earn R in interest.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

Reread the question and write down all of the important information that you know. This is a timeline question: look for where something changes (like an interest rate or a compounding period) to help you break the solution into smaller pieces. For example, in this question, the interest rate changes after 3 years.


STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first 3 years (from 2006 until 2009):

P=7 150i=8,96%=0,0896n=3A= ?

This is a compound interest problem:

A=P(1+i)n

Solve the first part of the problem:

A=7 150(1+0,0896365)(3×365)A=7 150(1,000...)1 095=7 150(1,308...)=R9 354,70The amount after 3 years.

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next 2 years (from 2009 until 2011):

P=R9 354,70i=8,33%=0,0833n=2A= ?

Now solve the second part of the problem:

A=9 354,7(1+0,0833365)(2×365)=9 354,7(1,000...)730=9 354,7(1,181...)=R11 050,33The amount after 2 years.

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next 4 years:

P=R11 050,33i=8,33%=0,0833n=4A= ?

Now solve the third part of the problem, the last 4 years:

A=11 050,33(1+0,083352)(4×52)=11 050,33(1,001...)208=11 050,33(1,395...)=R15 415,81The amount after 9 years.

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

Interest earned =15 415,817150=R8 265,81

In 2015:

  1. She will have R15 415,81 in her account.
  2. The amount of interest she earns is R8 265,81.

Submit your answer as: and

Birthday investment

When his son was 9 years old, Chike made a deposit of R3 550 in the bank. The investment grew at a simple interest rate and when Chike's son was 36 years old, the value of the investment was R8 917,60.

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a long word problem. Start by reading through it again and writing down the information you think you need to calculate the answer. Also write down the formula you think you need to answer the question.


STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=8 917,60
  • P=3 550,00
  • i=?
  • n=369=27

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make i the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1n=i

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

i=(8 917,603 550)127=0,056=5,6% per annum

The simple interest rate is 5,6% per annum.


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Timeline question: interest earned

Imagine that a friend of yours, Masoabi, deposits R3 500 into a bank account. After 11 years the account has a value of R9 020,54.

  1. What is the total amount of interest which the account accrues?
    Answer: The interest is R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The money in the account at the end come from two different sources: Masoabi, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Masoabi's original deposit.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The interest is the amount of money the bank pays into the account; it is different to the interest rate!

    The final amount of money in the account is made of the money Masoabi put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Masoabi put into the account herself.

    interest earned=9 020,543 500=5 520,54

    Masoabi's account earns R5 520,54 during these 11 years.


    Submit your answer as:
  2. Now imagine that Masoabi takes R2 500 out of the account 8 years after she opened the account. Then after 11 years her account has a final value of R5 784,04. Determine how much interest the bank paid into her account in this situation.
    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Unlike in the first question, the amount of Masoabi's money in the account changes - she takes money from the account. We must deal with this change now when we calculate the interest.

    The total amount of money Masoabi contributes to the account is the original deposit of R3 500 minus the R2 500 she took out of the account after 8 years. Therefore, in the end, Masoabi has put 3 5002 500=1 000 into the account.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we know that R1 000 of the R5 784,04 in the account came from Masoabi's pocket: the rest of it is interest paid into the account by the bank.

    So we must do what we did in the first question: subtract Masoabi's money from the amount in the account:

    interest earned=5 784,041 000=4 784,04

    There we have it: the bank pays R4 784,04 into the account.


    Submit your answer as:

Timelines: following money as it comes and goes

In 2002 Kamogelo opens a savings account at Vryburg National Beefeaters Bank which pays an interest rate of 9,76 % p.a. The interest is compounded every week. Kamogelo's original deposit is R5 800. Then in 2009 he withdraws R2 300 from the account. Later, in 2015, he takes R850 from the account.

  1. If the interest rate stays unchanged the entire time, how much money will be in the account in 2018?

    Answer: The total amount in the account will be R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.


    STEP: <no title>
    [−2 points ⇒ 4 / 6 points left]

    Determine what has been provided and what is required from 2002 to 2009:

    P=R5 800i=9,76%=0,0976n=20092002=7 yearsA= ?

    Solve the first part of the problem:

    A=5 800(1+0,097652)(7×52)A=5 800(1,001...)364=5 800(1,978...)=R11 477,83The amount after 7 years.

    STEP: <no title>
    [−2 points ⇒ 2 / 6 points left]

    Determine what has been provided and what is required for the next 6 years. Did Kamogelo put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

    P=11 477,832300=R9 177,83i=9,76%=0,0976n=20152009=6 yearsA= ?

    Now solve the second part of the problem, from 2009 to 2015:

    A=9 177,83(1+0,097652)(6×52)=9 177,83(1,001...)312=9 177,83(1,795...)=R16 474,96The amount after 13 years.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 3 years. Did Kamogelo put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

    P=16 474,96850=R15 624,96i=9,76%=0,0976n=20182015=3 yearsA= ?

    Now solve the third part of the problem, from 2 015 to 2 018:

    A=15 624,96(1+0,097652)(3×52)=15 624,96(1,001...)156=15 624,96(1,339...)=R20 934,43The amount after 16 years.

    After 16 years, Kamogelo will have R20 934,43 in his account.


    Submit your answer as:
  2. How much interest will the account earn during the entire 16 year period?

    Answer: The amount of interest he will get is R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Be careful here! Remember that when you calculate the value of A in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

    Interest from the first period=11 477,835 800,00=5 677,83Interest from the second period=16 474,969 177,83=7 297,13Interest from the third period=20 934,4315 624,96=5 309,47Total interest=R18 284,43

    (The total interest is the sum of the three different interest amounts!)

    The final answer is: After 16 years, the account earns R18 284,43 of interest.


    Submit your answer as:

Savings investment

A person invests an amount of R4 280 in a savings account which pays simple interest at a rate of 9% per annum.

Calculate the balance accumulated by the end of 7 years.

Answer: The final balance is: R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.


STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=4 280
  • i=9100
  • n=7

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

A=P(1+in)=R4 280(1+(9100)×7)=R6 976,40

The final balance in the account is R6 976,40.


Submit your answer as:

Saving for a big purchase

Abdulai wants to buy a large refrigerator, but right now he doesn't have enough money. A friend told Abdulai that in 7 years the large refrigerator will cost R9 150. He decides to start saving money today at Port Elizabeth First Bank. Abdulai deposits R4 900 into a savings account with an interest rate of 10,24% p.a. compounded quarterly. Then after 1,5 years the bank changes the interest rate to 10,23% p.a. compounded daily. After another 2,5 years, the interest rate changes again to 9,82% p.a. compounded monthly.

How much money will Abdulai have in the account after 7 years, and will he then have enough money to buy the large refrigerator?

Answer:

He will have R .

Will he have enough money?

numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]
This question is a beast! When you encounter big questions like this one, try to break the question into smaller pieces. Also, it might be helpful to draw a diagram (a timeline) of what is happening in this situation.
STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first 1,5 years:

P=R4 900i=10,24%=0,1024n=1,5 yearsA= ?
TIP: Remember to convert months to years if necessary!

Solve the first part of the problem:

A=4 900(1+0,10244)(1,5×4)=4 900(1,025...)6,0=4 900(1,163...)=R5 702,49

This is the amount after 1,5 years.


STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next 2,5 years:

P=R5 702,49i=10,23%=0,1023n=2,5 yearsA= ?
TIP: Don't forget to convert months to years if necessary!

Now solve the second part of the problem:

A=5 702,49(1+0,1023365)(2,5×365)=5 702,49(1,000...)912,5=5 702,49(1,291...)=R7 364,10

This is the amount after 2,5 years.


STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next 3 years:

P=R7 364,10i=9,82%=0,0982n=3A= ?

Now solve the third part of the problem, the last 3 years:

A=7 364,1(1+0,098212)(3×12)=7 364,1(1,008...)36=7 364,1(1,341...)=R9 875,12

This is the amount after 7 years.


STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After 7 years, Abdulai will have R9 875,12 in the account. Yes, he will be able to buy the large refrigerator.


Submit your answer as: and

Simple interest investment

Carl wants to invest R2 110 at a simple interest rate of 7,4% p.a.

How many years will it take for the money to grow to R8 018?

INSTRUCTION: Round up your answer to the nearest year.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.


STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

  • A=8 018
  • P=2 110
  • i=7,4100
  • n=?

STEP: Transform the simple interest formula for n
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make n the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1i=n

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

n=(8 0182 110)1(7,4100)=37,83783...=38 yearsround UP to the nearest integer

Rounding up to the nearest year, it will take 38 years to reach the goal of saving R8 018.

The answer, rounded up to the nearest integer, is 38 years.


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Compound interest rate

Babangida invests R1 640 into an account which pays out a lump sum at the end of 9 years.

If he gets R2 000,80 at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=2 000,80
  • P=1 640,00
  • i=?
  • n=9

STEP: Transform the compound interest formula for i
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make i the subject of the formula:

A=P(1+i)nAP=(1+i)n(AP)1n=1+i(AP)1n1=i

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

i=(2 000,801 640)191=0,022=2,2% per annum

For the situation in this question, the effective interest rate is 2,2% per annum.


Submit your answer as:

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

  • Nkosingiphile goes to We Pay More Bank and deposits R3 750,00 into an account which pays 7,7% p.a. compounded every six months. After 4 years the bank changes the interest rate to 6,2% p.a., still compounded every six months. Nkosingiphile leaves the money in the account for a total of 9 years.

  1. What change happened to the account, and in which year did it happen?
    Answer: What changes?
    The change happened after years.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.
    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    There is a lot of information in story about the investment. The key part for this question is: ' After 4 years the bank changes the interest rate .'

    During this investment, the interest percentage rate changed and it happens 4 years after Nkosingiphile opens the account


    Submit your answer as: and
  2. How many years passed from the change in the account to the final year?
    Answer: years
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The investment lasts a total of 9 years, and the change happens 4 years after Nkosingiphile opens the account. How many years are missing at the end?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The investment lasts a total of 9 years; and from above we know that there was a change 4 years after the account was opened. Then the number of years from the change to the end must be 94=5.

    It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 5 years at the end of the investment period.


    Submit your answer as:
  3. What was the interest rate in the 4th year of the investment?
    Answer: %
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    In the story, the bank makes a change to the interest rate. When does this happen? In the 4th year which interest information applies?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The interest rate for the account changes after 4 years; the 4th year is before the change occurs. Therefore the interest rate in the 4th year is the original rate, 7,7%.


    Submit your answer as:

Finding the length of an investment

Some years ago, a person opened a savings account at Limpopo West Bank. The principal amount was R3 500 but now the account has a value of R6 195,50. For how many years was the money in the account if the account gets 8,5% compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.
Answer: The money was in the account for years.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.


STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

A=P(1+i)nWhere: A=6 195,5P=3 500i=8,5%=0,085n=?

STEP: Substitute the values into the formula and begin solving for n
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of n. Substitute in the known values and make n the subject of the formula:

6 195,5=3 500(1+0,085)nthis equation by 3 500divide both side of6 195,5=3 500(1,085)n6 195,53 500=(1,085)n

STEP: Rewrite the equation in logarithmic form and calculate n
[−1 point ⇒ 0 / 3 points left]

To make n the subject of the formula, we convert the equation from exponential form to logarithmic form:

n=log1,085(6 195,53 500)n=log(1,7701...)log1,085

Use a calculator to evaluate the logarithm:

n=7

Therefore, the person left the money in the account for 7 years.


Submit your answer as:

Compound interest investment

Halima wants to invest some money at a compound interest rate of 12,0% p.a.

How much money should be invested if she wants to reach a sum of R63 400 in 6 years' time?

INSTRUCTION: Round up your answer to the nearest rand.
Answer: Halima must invest R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=R63 400
  • P=?
  • i=12,0%=12,0100
  • n=6

STEP: Transform the equation for P
[−1 point ⇒ 1 / 3 points left]

The question states that the investment grows at a compound interest rate. So we will use the compound interest formula. To determine the amount she must invest, we can to make P the subject of the formula:

A=P(1+i)nA(1+i)n=P

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]
63 400(1+12,0100)6=P63 400(1,12)6=PP=R32 120,41

The result of the calculation shows that Halima must invest R32 120,41 if she is going to reach her goal of R63 400 in 6 years. The question tells us to "round up your answer to the nearest rand," so the final answer is R32 121.

NOTE:

Rounding up is important here. In this question, Halima is aiming to reach a final goal of R63 400. If we round the answer down, then after 6 years the total amount of money will not quite reach R63 400 - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is R32 121.


Submit your answer as:

Timeline questions: changing interest rates

Ladi goes to Save Here Bank and deposits R4 000,00 into an account which pays 7,7% p.a. compounded two times per year. After 4 years the bank changes the interest rate to 9,0% p.a., still compounded two times per year.

  1. How much money will Ladi have in her account 9 years after the original investment?

    Answer: Total amount = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.


    STEP: <no title>
    [−3 points ⇒ 3 / 6 points left]

    This question describes a 9 year period which is broken into two sections because of the change in the interest rate: the first section lasts 4 years (while the interest rate is 7,7%); the second section lasts 5 years (when the rate is 9,0%).

    Determine what has been provided and what is required for the first 4 years:

    P=4 000i=7,7%=0,077n=4A= ?

    This is a compound interest problem:

    A=P(1+i)n

    Solve the first part of the problem. Remember that we must multiply the number of years by 2 and divide the interest rate by 2 because the interest is compounded two times per year.

    A=4 000,0(1+0,0772)(4×2)A=4 000,0(1,038...)8=4 000,0(1,352...)=R5 411,43

    This is the amount after 4 years.


    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 5 years. All of the money from the above calculation remains in the account - it is the starting point for the last 5 years.

    P=R5 411,43i=9,0%=0,09n=5A= ?

    Now solve the second part of the problem, the last 5 years (again, we must adjust the interest rate and the the number of years by 2.):

    A=5 411,43(1+0,092)(5×2)=5 411,43(1,045...)10=5 411,43(1,553...)=R8 403,79

    After 9 years, Ladi will have R8 403,79 in her account.


    Submit your answer as:
  2. What is the total amount of interest the bank pays into her account?

    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The amount of interest is what the bank paid into the account on top of what Ladi put into the account at the beginning.


    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    Use the answer from the work above to calculate how much interest the bank paid.

    Interest earned =R8 403,79R4 000,00=R4 403,79

    The amount of interest the bank pays is R4 403,79.


    Submit your answer as:

Compound interest investment

An amount of R1 930 is invested in a savings account which pays a compound interest rate of 6,4% p.a.

Calculate the balance accumulated by the end of 4 years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.


STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=1 930
  • i=6,4100
  • n=4

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

A=P(1+i)n=1 930(1+6,4100)4=1 930(1,064)4=R2 473,57

If you have an answer close to this, it is likely that you rounded off during the calculation. However, you should not round off until you reach the end of the calculation - let your calculator hold all of the decimal digits that come as you do the calculation until the end.

The balance after years is R2 473,57.


Submit your answer as:

Timelines: population growth

Farmer Marcel starts working with a beehive in his orchard. The beehive contains 61 000 bees. The number of bees in the hive grows by 2,32% each year. After 5 years farmer Marcel gives 9 000 of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate (2,32% every year).

  1. How many bees will be in farmer Marcel's hive 4 years after he gave some of his bees away?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in Marcel's hive (rounded to the nearest whole number).

    one-of
    type(numeric.abserror(0.5))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.


    STEP: Calculate the number of bees after 5 years
    [−2 points ⇒ 2 / 4 points left]

    Populations grow like compound interest, so we should use this formula: A=P(1+i)n.

    In this case we will use the equation as follows:

    • P is the number of bees at the beginning =61 000
    • i is the rate of growth of th hive =2,32%=0,0232
    • n is the number of years =5
    • A=?

    Solve the first part of the problem:

    A=61 000(1+0,0232)5=68 412,032...68 412
    TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So, there are approximately 68 412 bees in the hive after 5 years.


    STEP: Calculate the number of bees after 9 years
    [−2 points ⇒ 0 / 4 points left]

    At this point the farmer takes away (subtracts) 9 000 bees from his hive. You must subtract now because Marcel is taking bees out of the hive.

    P=68 4129 000=59 412i=2,32%=0,0232n=4A= ?

    Now calculate how many bees there will be after the last 4 years.

    A=59 412(1+0,0232)4=65 120,285...65 120

    So, after 9 years, farmer Marcel will have 65 120 bees in his beehive.

    NOTE:

    You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

    A=61 000(1+0,0232)99 000(1+0,0232)4=65 120,321...65 120

    This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.


    Submit your answer as:
  2. There were 9 000 in the second hive when it started. How many bees will be in the second hive after 4 years?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in the second hive (rounded to the nearest whole number).

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).


    STEP: Calculate the number of bees in the second hive after 4 years
    [−2 points ⇒ 0 / 2 points left]

    For this question, remember that the second hive starts with 9 000 bees, but it only starts 5 years after the first hive. Therefore the second hive only grows for 4 years.

    P=9 000i=2,32%=0,0232n=4A= ?
    A=9 000(1+0,0232)4=9 864,717...9 865

    So, after 4 years, the second hive had approximately 9 865 bees in it.


    Submit your answer as:
  3. A single bee can make about 0,56 grams of honey in a summer. If 90% of the bees in the colony produce honey, how much honey did the two hives produce in the ninth year after farmer Marcel started working with bees?

    INSTRUCTION: Write your answer in kilograms and then round it to two decimal places.
    Answer:

    The amount of honey produced is kg.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    How many bees are there in both hives after 9 years?


    STEP: Calculate the number of bees that produce honey
    [−1 point ⇒ 1 / 2 points left]

    90% of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate 90% (0,90) of the total number of bees:

    0,90×(65 120+9 865)=67 486,567 487
    TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So there are about 67 487 bees making honey.


    STEP: Calculate how much honey these bees make, by multiplying
    [−0 points ⇒ 1 / 2 points left]

    Each of the 67 487 bees produces 0,56 grams of honey. To find the total amount of honey, we just multiply:

    0,56×67 487=37 792,72

    STEP: Convert grams to kilograms
    [−1 point ⇒ 0 / 2 points left]

    But we aren't done yet because the question asks us to write the answer in kilograms. Every kg is equal to 1 000 g.

    The amount of honey (in kg) is:

    37 792,721000=37,7927237,79 kg

    The two hives produce approximately 37,79 kg of honey.


    Submit your answer as:

Exercises

Finding the length of an investment

At some time in the past, Mike deposited R2 750 into a savings account at The Bank of Money Money. The account accrues interest at a rate of 6,47% compounded yearly. How long ago did Mike open the account if the balance is now R3 442,68? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
What kind of interest was earned? Which formula do we need for this calculation?
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

A=P(1+i)nWhere: A=3 442,68P=2 750i=6,47%=0,0647n=?

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for n.

3 442,68=2 750(1+0,0647)n3 442,68=2 750(1,0647)n3 442,682 750=(1,0647)n

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that n is the subject of the formula:

n=log1,0647(3 442,682 750)n=3,58333...

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Mike left the money in the account for about 3,58 years. However, we need to give the answer in terms of years and months.

3,58333... years = 3 years and 0,58333... of a year (which will be in months).

We know that there are 12 months in a year. To convert 0,58333... of a year into months, we do the following calculation:

0,58333... of a year×(12 monthsyear)=7 month(s)

Therefore, Mike deposited the money into the account 3 years and 7 month(s) ago.


Submit your answer as: and

Finding the length of an investment

At some time in the past, Elethu deposited R2 450 into a savings account at The Bank of Money Money. The account accrues interest at a rate of 8,48% compounded yearly. How long ago did Elethu open the account if the balance is now R3 302,03? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
What kind of interest was earned? Which formula do we need for this calculation?
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

A=P(1+i)nWhere: A=3 302,03P=2 450i=8,48%=0,0848n=?

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for n.

3 302,03=2 450(1+0,0848)n3 302,03=2 450(1,0848)n3 302,032 450=(1,0848)n

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that n is the subject of the formula:

n=log1,0848(3 302,032 450)n=3,66666...

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Elethu left the money in the account for about 3,67 years. However, we need to give the answer in terms of years and months.

3,66666... years = 3 years and 0,66666... of a year (which will be in months).

We know that there are 12 months in a year. To convert 0,66666... of a year into months, we do the following calculation:

0,66666... of a year×(12 monthsyear)=8 month(s)

Therefore, Elethu deposited the money into the account 3 years and 8 month(s) ago.


Submit your answer as: and

Finding the length of an investment

At some time in the past, Litha deposited R2 100 into a savings account at The Bank of Money Money. The account accrues interest at a rate of 6,72% compounded yearly. How long ago did Litha open the account if the balance is now R3 119,23? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
What kind of interest was earned? Which formula do we need for this calculation?
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

A=P(1+i)nWhere: A=3 119,23P=2 100i=6,72%=0,0672n=?

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for n.

3 119,23=2 100(1+0,0672)n3 119,23=2 100(1,0672)n3 119,232 100=(1,0672)n

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that n is the subject of the formula:

n=log1,0672(3 119,232 100)n=6,08333...

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Litha left the money in the account for about 6,08 years. However, we need to give the answer in terms of years and months.

6,08333... years = 6 years and 0,08333... of a year (which will be in months).

We know that there are 12 months in a year. To convert 0,08333... of a year into months, we do the following calculation:

0,08333... of a year×(12 monthsyear)=1 month(s)

Therefore, Litha deposited the money into the account 6 years and 1 month(s) ago.


Submit your answer as: and

Timeline question: adding or withdrawing money

Pietermaritzburg Local Bank offers a savings account which receives 10,9% interest p.a. compounded semi-annually. Neels decides to open an account, and deposits R3 500. His account accrues interest for 5 years until he puts R2 500 more into the account. After that, Neels leaves the money in the account until 9 years after his original deposit. How much money will be in the account in the end?

Answer: Total amount = R
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.


STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of 9 years. This period is broken into two sections: 5 years after Neels opens his account, and then the last 4 years after he puts R2 500 more into his account. We must calculate the value in the account separately for these two sections of time.

For the first 5 years we have:

P=R3 500i=10,9%=0,109n=5A= ?

The question tells us that the interest rate is "10,9% interest p.a. compounded semi-annually" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by 2 and divide the interest rate by 2 because the interest compounds semi-annually.

A=P(1+i)nA=3 500(1+0,1092)(5×2)=3 500(1,054...)10=3 500(1,700...)=R5 950,23The amount after 5 years.

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Neels puts R2 500 more into his account: did Neels put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

P=5 950,23+2 500=R8 450,23i=10,9%=0,109n=4A= ?

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last 4 years. (As above, we must adjust the exponent and the interest rate.)

A=8 450,23(1+0,1092)(4×2)=8 450,23(1,054...)8=8 450,23(1,528...)=R12 919,37The amount after 9 years.

In the end, Neels will have R12 919,37 in his account at the bank.


Submit your answer as:

Timeline question: adding or withdrawing money

Northern Cape Provincial Bank offers a savings account which gets 7,6% interest p.a. compounded daily. Adebajo decides to open an account, and deposits R3 500. His account accrues interest for 8 years until he adds R2 500 more into the account. After that, Adebajo leaves the money in the account until 11 years after his original deposit. How much money will be in the account in the end?

Answer: Total amount = R
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.


STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of 11 years. This period is broken into two sections: 8 years after Adebajo opens his account, and then the last 3 years after he adds R2 500 more into his account. We must calculate the value in the account separately for these two sections of time.

For the first 8 years we have:

P=R3 500i=7,6%=0,076n=8A= ?

The question tells us that the interest rate is "7,6% interest p.a. compounded daily" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by 365 and divide the interest rate by 365 because the interest compounds daily.

A=P(1+i)nA=3 500(1+0,076365)(8×365)=3 500(1,000...)2 920=3 500(1,836...)=R6 428,23The amount after 8 years.

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Adebajo adds R2 500 more into his account: did Adebajo put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

P=6 428,23+2 500=R8 928,23i=7,6%=0,076n=3A= ?

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last 3 years. (As above, we must adjust the exponent and the interest rate.)

A=8 928,23(1+0,076365)(3×365)=8 928,23(1,000...)1 095=8 928,23(1,256...)=R11 214,35The amount after 11 years.

In the end, Adebajo will have R11 214,35 in his account at the bank.


Submit your answer as:

Timeline question: adding or withdrawing money

Mahikeng City Bank offers a savings account which receives 7,0% interest p.a. compounded two times each year. Mphikeleli decides to open an account, and deposits R6 500. His account accrues interest for 4 years until he adds R3 500 more into the account. After that, Mphikeleli leaves the money in the account until 10 years after his original deposit. How much money will be in the account in the end?

Answer: Total amount = R
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.


STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of 10 years. This period is broken into two sections: 4 years after Mphikeleli opens his account, and then the last 6 years after he adds R3 500 more into his account. We must calculate the value in the account separately for these two sections of time.

For the first 4 years we have:

P=R6 500i=7,0%=0,07n=4A= ?

The question tells us that the interest rate is "7,0% interest p.a. compounded two times each year" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by 2 and divide the interest rate by 2 because the interest compounds two times each year.

A=P(1+i)nA=6 500(1+0,072)(4×2)=6 500(1,035...)8=6 500(1,316...)=R8 559,26The amount after 4 years.

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Mphikeleli adds R3 500 more into his account: did Mphikeleli put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

P=8 559,26+3 500=R12 059,26i=7,0%=0,07n=6A= ?

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last 6 years. (As above, we must adjust the exponent and the interest rate.)

A=12 059,26(1+0,072)(6×2)=12 059,26(1,035...)12=12 059,26(1,511...)=R18 222,37The amount after 10 years.

In the end, Mphikeleli will have R18 222,37 in his account at the bank.


Submit your answer as:

Finding the length of an investment

Sive puts R800 into a bank account at the First Bank of Springbok. Sive's account earns interest at a rate of 5,77% p.a. compounded every month. After how many years will the bank account have a balance of R1 550,87?

INSTRUCTION: Give your answer to one decimal place.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable every month, as this affects i and n in the compound interest formula.


STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

A=P(1+i)nWhere: A=1 550,87P=800i=5,77%=0,0577n=?

STEP: Adjust the values of i and n based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded every month.

We must adjust the values of i and n according to the compounding period. For this question, we must make the following adjustments:

i0,057712 and n(n×12)

In this case, n represents the number of years and (n×12) represents the number of times the bank pays interest into the account.

1 550,87=800(1+0,057712)(n×12)

Simplify the equation:

1 550,87=800(1,0048...)12n1 550,87800=(1,0048...)12n

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that 12n becomes the subject of the formula:

12n=log1,00480833333(1 550,87800)12n=138

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for n:

12n=138n=13812n=11,5

Therefore, the money has been in the Sive's account for 11,5 years.

NOTE: If you got an answer close to this (for example, 11,62 years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is 11,5.


Submit your answer as:

Finding the length of an investment

Sive puts R400 into a bank account at the First Bank of Springbok. Sive's account accrues interest at a rate of 6,56% p.a. compounded quarterly. After how many years will the bank account have a balance of R553,80?

INSTRUCTION: Give your answer to one decimal place.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable quarterly, as this affects i and n in the compound interest formula.


STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

A=P(1+i)nWhere: A=553,8P=400i=6,56%=0,0656n=?

STEP: Adjust the values of i and n based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded quarterly.

We must adjust the values of i and n according to the compounding period. For this question, we must make the following adjustments:

i0,06564 and n(n×4)

In this case, n represents the number of years and (n×4) represents the number of times the bank pays interest into the account.

553,8=400(1+0,06564)(n×4)

Simplify the equation:

553,8=400(1,0164)4n553,8400=(1,0164)4n

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that 4n becomes the subject of the formula:

4n=log1,0164(553,8400)4n=20

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for n:

4n=20n=204n=5

Therefore, the money has been in the Sive's account for 5 years.

NOTE: If you got an answer close to this (for example, 5,12 years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is 5.


Submit your answer as:

Finding the length of an investment

Simosethu puts R600 into a savings account at the First Bank of Springbok. Simosethu's account accrues interest at a rate of 6,34% p.a. compounded four times per year. After how many years will the savings account have a balance of R848,01?

INSTRUCTION: Give your answer to one decimal place.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable four times per year, as this affects i and n in the compound interest formula.


STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

A=P(1+i)nWhere: A=848,01P=600i=6,34%=0,0634n=?

STEP: Adjust the values of i and n based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded four times per year.

We must adjust the values of i and n according to the compounding period. For this question, we must make the following adjustments:

i0,06344 and n(n×4)

In this case, n represents the number of years and (n×4) represents the number of times the bank pays interest into the account.

848,01=600(1+0,06344)(n×4)

Simplify the equation:

848,01=600(1,01584...)4n848,01600=(1,01584...)4n

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that 4n becomes the subject of the formula:

4n=log1,01585(848,01600)4n=22

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for n:

4n=22n=224n=5,5

Therefore, the money has been in the Simosethu's account for 5,5 years.

NOTE: If you got an answer close to this (for example, 5,62 years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is 5,5.


Submit your answer as:

Timelines

Tshegofatso wants to buy a motorcycle. The cost of the motorcycle is R55 500. In 2004 Tshegofatso opened an account at Pinelands Trust Bank with R23 000. Then in 2010 she took R2 000 from the account. In 2013 Tshegofatso made another change: she deposited R2 500 into the account. If the account pays 6,44% p.a. compounded quarterly, how much money does she have now, and is it enough to buy the motorcycle?

Answer:
  1. The total amount in the account now is R .
  2. Does she have enough money?
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2004 to 2010 is a period of 3 years.


STEP: Find the amount of money there was in 2010 (starting in 2004)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2004 to 2010:

P=R23 000i=6,44%=0,0644n=20102004=6 yearsA= ?

Solve the first part of the problem:

A=23 000(1+0,06444)(6×4)A=23 000(1,0161)24=23 000(1,46715...)=R33 744,47The amount in 2010.

STEP: Adjust the amount of money to reflect Tshegofatso's withdrawal from the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2010 to 2013. You must adjust the amount of money in the account if Tshegofatso put more money in or took money out.

P=33 744,472000=R31 744,47i=6,44%=0,0644n=20132010=3 yearsA= ?

Now solve this part of the problem:

A=31 744,47(1+0,06444)(3×4)=31 744,47(1,0161)12=31 744,47(1,21126...)=R38 450,81The amount in 2013.

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2013 to 2018: did Tshegofatso put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

P=38 450,81+2500=R40 950,81i=6,44%=0,0644n=20182013=5 yearsA= ?

Now solve the third part of the problem, from 2013 to 2018:

A=40 950,81(1+0,06444)(5×4)=40 950,81(1,0161)20=40 950,81(1,37635...)=R56 362,67The amount this year.

STEP: State the final answer (including whether or not Tshegofatso has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Tshegofatso to buy the motorcycle.

The answers are:

  1. This year Tshegofatso has R56 362,67 in her account, and so
  2. yes, she can afford the motorcycle.

Submit your answer as: and

Timelines

Ismail wants to buy a motorcycle. The cost of the motorcycle is R48 500. In 2004 Ismail opened an account at Maseru Money Bank with R17 000. Then in 2008 he put R3 750 more into the account. In 2014 Ismail made another change: he withdrew R3 000 from the account. If the account earns 8,32% p.a. compounded every six months, how much money does he have now, and is it enough to buy the motorcycle?

Answer:
  1. The total amount in the account now is R .
  2. Does he have enough money?
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2004 to 2008 is a period of 3 years.


STEP: Find the amount of money there was in 2008 (starting in 2004)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2004 to 2008:

P=R17 000i=8,32%=0,0832n=20082004=4 yearsA= ?

Solve the first part of the problem:

A=17 000(1+0,08322)(4×2)A=17 000(1,0416)8=17 000(1,38550...)=R23 553,57The amount in 2008.

STEP: Adjust the amount of money to reflect Ismail's deposit into the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2008 to 2014. You must adjust the amount of money in the account if Ismail put more money in or took money out.

P=23 553,57+3750=R27 303,57i=8,32%=0,0832n=20142008=6 yearsA= ?

Now solve this part of the problem:

A=27 303,57(1+0,08322)(6×2)=27 303,57(1,0416)12=27 303,57(1,63084...)=R44 527,78The amount in 2014.

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2014 to 2018: did Ismail put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

P=44 527,783000=R41 527,78i=8,32%=0,0832n=20182014=4 yearsA= ?

Now solve the third part of the problem, from 2014 to 2018:

A=41 527,78(1+0,08322)(4×2)=41 527,78(1,0416)8=41 527,78(1,38550...)=R57 536,90The amount this year.

STEP: State the final answer (including whether or not Ismail has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Ismail to buy the motorcycle.

The answers are:

  1. This year Ismail has R57 536,90 in his account, and so
  2. yes, he can afford the motorcycle.

Submit your answer as: and

Timelines

Xolile wants to buy a motorcycle. The cost of the motorcycle is R56 250. In 2001 Xolile opened an account at First Rand Bank with R22 000. Then in 2007 she took R2 500 from the account. In 2012 Xolile made another change: she took R3 100 from the account. If the account pays 8,22% p.a. compounded once each month, how much money does she have now, and is it enough to buy the motorcycle?

Answer:
  1. The total amount in the account now is R .
  2. Does she have enough money?
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2001 to 2007 is a period of 3 years.


STEP: Find the amount of money there was in 2007 (starting in 2001)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2001 to 2007:

P=R22 000i=8,22%=0,0822n=20072001=6 yearsA= ?

Solve the first part of the problem:

A=22 000(1+0,082212)(6×12)A=22 000(1,00685)72=22 000(1,63479...)=R35 965,53The amount in 2007.

STEP: Adjust the amount of money to reflect Xolile's withdrawal from the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2007 to 2012. You must adjust the amount of money in the account if Xolile put more money in or took money out.

P=35 965,532500=R33 465,53i=8,22%=0,0822n=20122007=5 yearsA= ?

Now solve this part of the problem:

A=33 465,53(1+0,082212)(5×12)=33 465,53(1,00685)60=33 465,53(1,50621...)=R50 406,22The amount in 2012.

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2012 to 2018: did Xolile put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

P=50 406,223100=R47 306,22i=8,22%=0,0822n=20182012=6 yearsA= ?

Now solve the third part of the problem, from 2012 to 2018:

A=47 306,22(1+0,082212)(6×12)=47 306,22(1,00685)72=47 306,22(1,63479...)=R77 336,05The amount this year.

STEP: State the final answer (including whether or not Xolile has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Xolile to buy the motorcycle.

The answers are:

  1. This year Xolile has R77 336,05 in her account, and so
  2. yes, she can afford the motorcycle.

Submit your answer as: and

Timeline questions: following the changes

At the beginning of 2005 a young woman starts a savings account at Mthatha Bank. She invests R8 650 into the account. The interest rate is 6,33 % p.a. compounded monthly. In 2010 the bank changes the interest rate to 6,33 % p.a. compounded semi-annually. Then in 2016, the interest rate changes again to 7,02 % p.a. compounded semi-annually.

  1. How much money will the woman have in her account in 2018, 13 years after her original deposit?
  2. How much interest will she earn from the bank during this time?
INSTRUCTION: Assume that every year has 365 days (ignore leap years).
Answer:
  1. She will have R in the account.
  2. She will earn R in interest.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

Reread the question and write down all of the important information that you know. This is a timeline question: look for where something changes (like an interest rate or a compounding period) to help you break the solution into smaller pieces. For example, in this question, the interest rate changes after 5 years.


STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first 5 years (from 2005 until 2010):

P=8 650i=6,33%=0,0633n=5A= ?

This is a compound interest problem:

A=P(1+i)n

Solve the first part of the problem:

A=8 650(1+0,063312)(5×12)A=8 650(1,005...)60=8 650(1,371...)=R11 860,67The amount after 5 years.

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next 6 years (from 2010 until 2016):

P=R11 860,67i=6,33%=0,0633n=6A= ?

Now solve the second part of the problem:

A=11 860,67(1+0,06332)(6×2)=11 860,67(1,031...)12=11 860,67(1,453...)=R17 238,43The amount after 6 years.

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next 2 years:

P=R17 238,43i=7,02%=0,0702n=2A= ?

Now solve the third part of the problem, the last 2 years:

A=17 238,43(1+0,07022)(2×2)=17 238,43(1,035...)4=17 238,43(1,148...)=R19 789,14The amount after 13 years.

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

Interest earned =19 789,148650=R11 139,14

In 2018:

  1. She will have R19 789,14 in her account.
  2. The amount of interest she earns is R11 139,14.

Submit your answer as: and

Timeline questions: following the changes

At the beginning of 2007 a young woman starts a savings account at Cape Town Bank. She deposits R5 950 into the account. The interest rate is 8,03 % p.a. compounded monthly. In 2012 the bank changes the interest rate to 8,03 % p.a. compounded semi-annually. Then in 2015, the interest rate changes again to 8,56 % p.a. compounded semi-annually.

  1. How much money will the woman have in her account in 2019, 12 years after her original deposit?
  2. How much interest will she earn from the bank during this time?
INSTRUCTION: Assume that every year has 365 days (ignore leap years).
Answer:
  1. She will have R in the account.
  2. She will earn R in interest.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

Reread the question and write down all of the important information that you know. This is a timeline question: look for where something changes (like an interest rate or a compounding period) to help you break the solution into smaller pieces. For example, in this question, the interest rate changes after 5 years.


STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first 5 years (from 2007 until 2012):

P=5 950i=8,03%=0,0803n=5A= ?

This is a compound interest problem:

A=P(1+i)n

Solve the first part of the problem:

A=5 950(1+0,080312)(5×12)A=5 950(1,006...)60=5 950(1,492...)=R8 877,80The amount after 5 years.

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next 3 years (from 2012 until 2015):

P=R8 877,80i=8,03%=0,0803n=3A= ?

Now solve the second part of the problem:

A=8 877,8(1+0,08032)(3×2)=8 877,8(1,040...)6=8 877,8(1,266...)=R11 242,97The amount after 3 years.

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next 4 years:

P=R11 242,97i=8,56%=0,0856n=4A= ?

Now solve the third part of the problem, the last 4 years:

A=11 242,97(1+0,08562)(4×2)=11 242,97(1,042...)8=11 242,97(1,398...)=R15 721,33The amount after 12 years.

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

Interest earned =15 721,335950=R9 771,33

In 2019:

  1. She will have R15 721,33 in her account.
  2. The amount of interest she earns is R9 771,33.

Submit your answer as: and

Timeline questions: following the changes

At the beginning of 2010 a young woman starts a savings account at Egoli Bank. She deposits R8 350 into the account. The interest rate is 9,73 % p.a. compounded monthly. In 2015 the bank changes the interest rate to 9,73 % p.a. compounded semi-annually. Then in 2021, the interest rate changes again to 8,94 % p.a. compounded semi-annually.

  1. How much money will the woman have in her account in 2023, 13 years after her original deposit?
  2. How much interest will she earn from the bank during this time?
INSTRUCTION: Assume that every year has 365 days (ignore leap years).
Answer:
  1. She will have R in the account.
  2. She will earn R in interest.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

Reread the question and write down all of the important information that you know. This is a timeline question: look for where something changes (like an interest rate or a compounding period) to help you break the solution into smaller pieces. For example, in this question, the interest rate changes after 5 years.


STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first 5 years (from 2010 until 2015):

P=8 350i=9,73%=0,0973n=5A= ?

This is a compound interest problem:

A=P(1+i)n

Solve the first part of the problem:

A=8 350(1+0,097312)(5×12)A=8 350(1,008...)60=8 350(1,623...)=R13 555,60The amount after 5 years.

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next 6 years (from 2015 until 2021):

P=R13 555,60i=9,73%=0,0973n=6A= ?

Now solve the second part of the problem:

A=13 555,6(1+0,09732)(6×2)=13 555,6(1,048...)12=13 555,6(1,768...)=R23 970,96The amount after 6 years.

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next 2 years:

P=R23 970,96i=8,94%=0,0894n=2A= ?

Now solve the third part of the problem, the last 2 years:

A=23 970,96(1+0,08942)(2×2)=23 970,96(1,044...)4=23 970,96(1,191...)=R28 553,00The amount after 13 years.

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

Interest earned =28 553,08350=R20 203,00

In 2023:

  1. She will have R28 553,00 in her account.
  2. The amount of interest she earns is R20 203,00.

Submit your answer as: and

Birthday investment

When his son was 9 years old, Lefu made a deposit of R2 080 in the bank. The investment grew at a simple interest rate and when Lefu's son was 45 years old, the value of the investment was R5 898,88.

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a long word problem. Start by reading through it again and writing down the information you think you need to calculate the answer. Also write down the formula you think you need to answer the question.


STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=5 898,88
  • P=2 080,00
  • i=?
  • n=459=36

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make i the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1n=i

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

i=(5 898,882 080)136=0,051=5,1% per annum

The simple interest rate is 5,1% per annum.


Submit your answer as:

Birthday investment

When his son was 4 years old, Neels made a deposit of R6 500 in the bank. The investment grew at a simple interest rate and when Neels's son was 28 years old, the value of the investment was R13 520,00.

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a long word problem. Start by reading through it again and writing down the information you think you need to calculate the answer. Also write down the formula you think you need to answer the question.


STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=13 520,00
  • P=6 500,00
  • i=?
  • n=284=24

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make i the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1n=i

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

i=(13 520,006 500)124=0,045=4,5% per annum

The simple interest rate is 4,5% per annum.


Submit your answer as:

Birthday investment

When his son was 8 years old, Eric made a deposit of R5 500 in the bank. The investment grew at a simple interest rate and when Eric's son was 32 years old, the value of the investment was R11 836,00.

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a long word problem. Start by reading through it again and writing down the information you think you need to calculate the answer. Also write down the formula you think you need to answer the question.


STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=11 836,00
  • P=5 500,00
  • i=?
  • n=328=24

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make i the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1n=i

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

i=(11 836,005 500)124=0,048=4,8% per annum

The simple interest rate is 4,8% per annum.


Submit your answer as:

Timeline question: interest earned

Imagine that a friend of yours, Carina, deposits R6 500 into a bank account. After 11 years the account has a value of R16 586,22.

  1. What is the total amount of interest which the account accrues?
    Answer: The interest is R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The money in the account at the end come from two different sources: Carina, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Carina's original deposit.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The interest is the amount of money the bank pays into the account; it is different to the interest rate!

    The final amount of money in the account is made of the money Carina put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Carina put into the account herself.

    interest earned=16 586,226 500=10 086,22

    Carina's account earns R10 086,22 during these 11 years.


    Submit your answer as:
  2. Now imagine that Carina takes R4 500 out of the account 7 years after she opened the account. Then after 11 years her account has a final value of R10 259,88. Determine how much interest the bank paid into her account in this situation.
    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Unlike in the first question, the amount of Carina's money in the account changes - she takes money from the account. We must deal with this change now when we calculate the interest.

    The total amount of money Carina contributes to the account is the original deposit of R6 500 minus the R4 500 she took out of the account after 7 years. Therefore, in the end, Carina has put 6 5004 500=2 000 into the account.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we know that R2 000 of the R10 259,88 in the account came from Carina's pocket: the rest of it is interest paid into the account by the bank.

    So we must do what we did in the first question: subtract Carina's money from the amount in the account:

    interest earned=10 259,882 000=8 259,88

    There we have it: the bank pays R8 259,88 into the account.


    Submit your answer as:

Timeline question: interest earned

Imagine that a friend of yours, Lisa, deposits R6 000 into a bank account. After 10 years the account has a value of R11 838,01.

  1. What is the total amount of interest the bank pays into her account?
    Answer: The interest is R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The money in the account at the end come from two different sources: Lisa, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Lisa's original deposit.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The interest is the amount of money the bank pays into the account; it is different to the interest rate!

    The final amount of money in the account is made of the money Lisa put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Lisa put into the account herself.

    interest earned=11 838,016 000=5 838,01

    Lisa's account earns R5 838,01 during these 10 years.


    Submit your answer as:
  2. Now imagine that Lisa deposits R2 500 more into the account 4 years after she opened the account. Then after 10 years her account has a final value of R15 596,52. Determine how much interest the bank paid into her account in this situation.
    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Unlike in the first question, the amount of Lisa's money in the account changes - she deposits more money into the account. We must deal with this change now when we calculate the interest.

    The total amount of money Lisa contributes to the account is the original deposit of R6 000 plus the R2 500 she added to the account after 4 years. Therefore, in the end, Lisa has put 6 000+2 500=8 500 into the account.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we know that R8 500 of the R15 596,52 in the account came from Lisa's pocket: the rest of it is interest paid into the account by the bank.

    So we must do what we did in the first question: subtract Lisa's money from the amount in the account:

    interest earned=15 596,528 500=7 096,52

    There we have it: the bank pays R7 096,52 into the account.


    Submit your answer as:

Timeline question: interest earned

Imagine that a friend of yours, Amarachi, deposits R5 500 into a bank account. After 6 years the account has a value of R9 327,35.

  1. What is the total amount of interest the bank pays into her account?
    Answer: The interest is R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The money in the account at the end come from two different sources: Amarachi, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Amarachi's original deposit.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The interest is the amount of money the bank pays into the account; it is different to the interest rate!

    The final amount of money in the account is made of the money Amarachi put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Amarachi put into the account herself.

    interest earned=9 327,355 500=3 827,35

    Amarachi's account earns R3 827,35 during these 6 years.


    Submit your answer as:
  2. Now imagine that Amarachi puts R3 000 more into the account 4 years after she opened the account. Then after 6 years her account has a final value of R12 904,90. Determine how much interest the bank paid into her account in this situation.
    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Unlike in the first question, the amount of Amarachi's money in the account changes - she puts more money into the account. We must deal with this change now when we calculate the interest.

    The total amount of money Amarachi contributes to the account is the original deposit of R5 500 plus the R3 000 she added to the account after 4 years. Therefore, in the end, Amarachi has put 5 500+3 000=8 500 into the account.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we know that R8 500 of the R12 904,90 in the account came from Amarachi's pocket: the rest of it is interest paid into the account by the bank.

    So we must do what we did in the first question: subtract Amarachi's money from the amount in the account:

    interest earned=12 904,98 500=4 404,9

    There we have it: the bank pays R4 404,90 into the account.


    Submit your answer as:

Timelines: following money as it comes and goes

In 2007 Mphikeleli opens a savings account at Vryburg National Beefeaters Bank which receives an interest rate of 10,06 % p.a. The interest is compounded daily. Mphikeleli's original deposit is R6 600. Then in 2010 he deposits R2 800 more into the account. Later, in 2015, he takes R825 from the account.

  1. If the interest rate stays unchanged the entire time, how much money will be in the account in 2018?

    Answer: The total amount in the account will be R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.


    STEP: <no title>
    [−2 points ⇒ 4 / 6 points left]

    Determine what has been provided and what is required from 2007 to 2010:

    P=R6 600i=10,06%=0,1006n=20102007=3 yearsA= ?

    Solve the first part of the problem:

    A=6 600(1+0,1006365)(3×365)A=6 600(1,000...)1 095=6 600(1,352...)=R8 924,75The amount after 3 years.

    STEP: <no title>
    [−2 points ⇒ 2 / 6 points left]

    Determine what has been provided and what is required for the next 5 years. Did Mphikeleli put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

    P=8 924,75+2800=R11 724,75i=10,06%=0,1006n=20152010=5 yearsA= ?

    Now solve the second part of the problem, from 2010 to 2015:

    A=11 724,75(1+0,1006365)(5×365)=11 724,75(1,000...)1 825=11 724,75(1,653...)=R19 387,58The amount after 8 years.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 3 years. Did Mphikeleli put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

    P=19 387,58825=R18 562,58i=10,06%=0,1006n=20182015=3 yearsA= ?

    Now solve the third part of the problem, from 2 015 to 2 018:

    A=18 562,58(1+0,1006365)(3×365)=18 562,58(1,000...)1 095=18 562,58(1,352...)=R25 100,96The amount after 11 years.

    After 11 years, Mphikeleli will have R25 100,96 in his account.


    Submit your answer as:
  2. How much interest will the account earn during the entire 11 year period?

    Answer: The amount of interest he will get is R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Be careful here! Remember that when you calculate the value of A in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

    Interest from the first period=8 924,756 600,00=2 324,75Interest from the second period=19 387,5811 724,75=7 662,83Interest from the third period=25 100,9618 562,58=6 538,38Total interest=R16 525,96

    (The total interest is the sum of the three different interest amounts!)

    The final answer is: After 11 years, the account earns R16 525,96 of interest.


    Submit your answer as:

Timelines: following money as it comes and goes

In 2004 Werner opens a savings account at Polokwane Local Bank which gets an interest rate of 9,28 % p.a. The interest is compounded daily. Werner's original deposit is R5 950. Then in 2008 he puts R2 100 more into the account. Later, in 2013, he takes R925 from the account.

  1. If the interest rate stays unchanged the entire time, how much money will be in the account in 2015?

    Answer: The total amount in the account will be R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.


    STEP: <no title>
    [−2 points ⇒ 4 / 6 points left]

    Determine what has been provided and what is required from 2004 to 2008:

    P=R5 950i=9,28%=0,0928n=20082004=4 yearsA= ?

    Solve the first part of the problem:

    A=5 950(1+0,0928365)(4×365)A=5 950(1,000...)1 460=5 950(1,449...)=R8 623,96The amount after 4 years.

    STEP: <no title>
    [−2 points ⇒ 2 / 6 points left]

    Determine what has been provided and what is required for the next 5 years. Did Werner put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

    P=8 623,96+2100=R10 723,96i=9,28%=0,0928n=20132008=5 yearsA= ?

    Now solve the second part of the problem, from 2008 to 2013:

    A=10 723,96(1+0,0928365)(5×365)=10 723,96(1,000...)1 825=10 723,96(1,590...)=R17 054,63The amount after 9 years.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 2 years. Did Werner put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

    P=17 054,63925=R16 129,63i=9,28%=0,0928n=20152013=2 yearsA= ?

    Now solve the third part of the problem, from 2 013 to 2 015:

    A=16 129,63(1+0,0928365)(2×365)=16 129,63(1,000...)730=16 129,63(1,203...)=R19 418,66The amount after 11 years.

    After 11 years, Werner will have R19 418,66 in his account.


    Submit your answer as:
  2. How much interest will the account earn during the entire 11 year period?

    Answer: The amount of interest he will get is R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Be careful here! Remember that when you calculate the value of A in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

    Interest from the first period=8 623,965 950,00=2 673,96Interest from the second period=17 054,6310 723,96=6 330,67Interest from the third period=19 418,6616 129,63=3 289,03Total interest=R12 293,66

    (The total interest is the sum of the three different interest amounts!)

    The final answer is: After 11 years, the account earns R12 293,66 of interest.


    Submit your answer as:

Timelines: following money as it comes and goes

In 2002 Chizoba opens a savings account at Masambe Bank which earns an interest rate of 10,37 % p.a. The interest is compounded monthly. Chizoba's original deposit is R6 500. Then in 2008 he adds R2 800 more into the account. Later, in 2013, he takes R1 075 from the account.

  1. If the interest rate stays unchanged the entire time, how much money will be in the account in 2017?

    Answer: The total amount in the account will be R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.


    STEP: <no title>
    [−2 points ⇒ 4 / 6 points left]

    Determine what has been provided and what is required from 2002 to 2008:

    P=R6 500i=10,37%=0,1037n=20082002=6 yearsA= ?

    Solve the first part of the problem:

    A=6 500(1+0,103712)(6×12)A=6 500(1,008...)72=6 500(1,858...)=R12 077,32The amount after 6 years.

    STEP: <no title>
    [−2 points ⇒ 2 / 6 points left]

    Determine what has been provided and what is required for the next 5 years. Did Chizoba put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

    P=12 077,32+2800=R14 877,32i=10,37%=0,1037n=20132008=5 yearsA= ?

    Now solve the second part of the problem, from 2008 to 2013:

    A=14 877,32(1+0,103712)(5×12)=14 877,32(1,008...)60=14 877,32(1,675...)=R24 930,96The amount after 11 years.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 4 years. Did Chizoba put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

    P=24 930,961075=R23 855,96i=10,37%=0,1037n=20172013=4 yearsA= ?

    Now solve the third part of the problem, from 2 013 to 2 017:

    A=23 855,96(1+0,103712)(4×12)=23 855,96(1,008...)48=23 855,96(1,511...)=R36 055,23The amount after 15 years.

    After 15 years, Chizoba will have R36 055,23 in his account.


    Submit your answer as:
  2. How much interest will the account earn during the entire 15 year period?

    Answer: The amount of interest he will get is R .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Be careful here! Remember that when you calculate the value of A in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

    Interest from the first period=12 077,326 500,00=5 577,32Interest from the second period=24 930,9614 877,32=10 053,64Interest from the third period=36 055,2323 855,96=12 199,27Total interest=R27 830,23

    (The total interest is the sum of the three different interest amounts!)

    The final answer is: After 15 years, the account earns R27 830,23 of interest.


    Submit your answer as:

Savings investment

A person invests an amount of R4 130 in a savings account which pays simple interest at a rate of 9% per annum.

Calculate the balance accumulated by the end of 3 years.

Answer: The final balance is: R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.


STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=4 130
  • i=9100
  • n=3

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

A=P(1+in)=R4 130(1+(9100)×3)=R5 245,10

The final balance in the account is R5 245,10.


Submit your answer as:

Savings investment

A person invests an amount of R5 980 in a savings account which pays simple interest at a rate of 7% per annum.

Calculate the balance accumulated by the end of 3 years.

Answer: The final balance is: R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.


STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=5 980
  • i=7100
  • n=3

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

A=P(1+in)=R5 980(1+(7100)×3)=R7 235,80

The final balance in the account is R7 235,80.


Submit your answer as:

Savings investment

A person invests an amount of R1 990 in a savings account which pays simple interest at a rate of 11% per annum.

Calculate the balance accumulated by the end of 5 years.

Answer: The final balance is: R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.


STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=1 990
  • i=11100
  • n=5

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

A=P(1+in)=R1 990(1+(11100)×5)=R3 084,50

The final balance in the account is R3 084,50.


Submit your answer as:

Saving for a big purchase

Lisa wants to buy a computer, but right now she doesn't have enough money. A friend told Lisa that in 8 years the computer will cost R9 250. She decides to start saving money today at Egoli Bank. Lisa saves R5 875 into a savings account with an interest rate of 5,82% p.a. compounded every 3 months. Then after 1,5 years the bank changes the interest rate to 5,8% p.a. compounded semi-annually. After another 2,5 years, the interest rate changes again to 5,01% p.a. compounded weekly.

How much money will Lisa have in the account after 8 years, and will she then have enough money to buy the computer?

Answer:

She will have R .

Will she have enough money?

numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]
This question is a beast! When you encounter big questions like this one, try to break the question into smaller pieces. Also, it might be helpful to draw a diagram (a timeline) of what is happening in this situation.
STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first 1,5 years:

P=R5 875i=5,82%=0,0582n=1,5 yearsA= ?
TIP: Remember to convert months to years if necessary!

Solve the first part of the problem:

A=5 875(1+0,05824)(1,5×4)=5 875(1,014...)6,0=5 875(1,090...)=R6 406,91

This is the amount after 1,5 years.


STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next 2,5 years:

P=R6 406,91i=5,8%=0,058n=2,5 yearsA= ?
TIP: Don't forget to convert months to years if necessary!

Now solve the second part of the problem:

A=6 406,91(1+0,0582)(2,5×2)=6 406,91(1,029...)5,0=6 406,91(1,153...)=R7 391,38

This is the amount after 2,5 years.


STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next 4 years:

P=R7 391,38i=5,01%=0,0501n=4A= ?

Now solve the third part of the problem, the last 4 years:

A=7 391,38(1+0,050152)(4×52)=7 391,38(1,001...)208=7 391,38(1,221...)=R9 030,59

This is the amount after 8 years.


STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After 8 years, Lisa will have R9 030,59 in the account. No, she will not be able to buy the computer.


Submit your answer as: and

Saving for a big purchase

Peter wants to buy a computer, but right now he doesn't have enough money. A friend told Peter that in 6 years the computer will cost R9 050. He decides to start saving money today at Durban United Bank. Peter saves R5 775 into a savings account with an interest rate of 9,2% p.a. compounded every 3 months. Then after 30 months the bank changes the interest rate to 7,85% p.a. compounded every two months. After another 6 months, the interest rate changes again to 7,65% p.a. compounded every 3 months.

How much money will Peter have in the account after 6 years, and will he then have enough money to buy the computer?

Answer:

He will have R .

Will he have enough money?

numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]
This question is a beast! When you encounter big questions like this one, try to break the question into smaller pieces. Also, it might be helpful to draw a diagram (a timeline) of what is happening in this situation.
STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first 30 months:

P=R5 775i=9,2%=0,092n=30 monthsA= ?
TIP: Remember to convert months to years if necessary!

Solve the first part of the problem:

A=5 775(1+0,0924)(2,5×4)=5 775(1,023...)10,0=5 775(1,255...)=R7 249,50

This is the amount after 30 months.


STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next 6 months:

P=R7 249,50i=7,85%=0,0785n=6 monthsA= ?
TIP: Don't forget to convert months to years if necessary!

Now solve the second part of the problem:

A=7 249,5(1+0,07856)(0,5×6)=7 249,5(1,013...)3,0=7 249,5(1,039...)=R7 537,78

This is the amount after 6 months.


STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next 3 years:

P=R7 537,78i=7,65%=0,0765n=3A= ?

Now solve the third part of the problem, the last 3 years:

A=7 537,78(1+0,07654)(3×4)=7 537,78(1,019...)12=7 537,78(1,255...)=R9 461,78

This is the amount after 6 years.


STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After 6 years, Peter will have R9 461,78 in the account. Yes, he will be able to buy the computer.


Submit your answer as: and

Saving for a big purchase

Safiyah wants to buy a huge television, but right now she doesn't have enough money. A friend told Safiyah that in 6 years the huge television will cost R8 850. She decides to start saving money today at Limpopo Trust Bank. Safiyah invests R4 950 into a savings account with an interest rate of 7,23% p.a. compounded two times per year. Then after 18 months the bank changes the interest rate to 5,85% p.a. compounded quarterly. After another 6 months, the interest rate changes again to 5,52% p.a. compounded semi-annually.

How much money will Safiyah have in the account after 6 years, and will she then have enough money to buy the huge television?

Answer:

She will have R .

Will she have enough money?

numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]
This question is a beast! When you encounter big questions like this one, try to break the question into smaller pieces. Also, it might be helpful to draw a diagram (a timeline) of what is happening in this situation.
STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first 18 months:

P=R4 950i=7,23%=0,0723n=18 monthsA= ?
TIP: Remember to convert months to years if necessary!

Solve the first part of the problem:

A=4 950(1+0,07232)(1,5×2)=4 950(1,036...)3,0=4 950(1,112...)=R5 506,47

This is the amount after 18 months.


STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next 6 months:

P=R5 506,47i=5,85%=0,0585n=6 monthsA= ?
TIP: Don't forget to convert months to years if necessary!

Now solve the second part of the problem:

A=5 506,47(1+0,05854)(0,5×4)=5 506,47(1,014...)2,0=5 506,47(1,029...)=R5 668,71

This is the amount after 6 months.


STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next 4 years:

P=R5 668,71i=5,52%=0,0552n=4A= ?

Now solve the third part of the problem, the last 4 years:

A=5 668,71(1+0,05522)(4×2)=5 668,71(1,027...)8=5 668,71(1,243...)=R7 048,18

This is the amount after 6 years.


STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After 6 years, Safiyah will have R7 048,18 in the account. No, she will not be able to buy the huge television.


Submit your answer as: and

Simple interest investment

Delphino wants to invest R2 310 at a simple interest rate of 8,4% p.a.

How many years will it take for the money to grow to R9 933?

INSTRUCTION: Round up your answer to the nearest year.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.


STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

  • A=9 933
  • P=2 310
  • i=8,4100
  • n=?

STEP: Transform the simple interest formula for n
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make n the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1i=n

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

n=(9 9332 310)1(8,4100)=39,28571...=40 yearsround UP to the nearest integer

Rounding up to the nearest year, it will take 40 years to reach the goal of saving R9 933.

The answer, rounded up to the nearest integer, is 40 years.


Submit your answer as:

Simple interest investment

Adedapo wants to invest R3 590 at a simple interest rate of 9,0% p.a.

How many years will it take for the money to grow to R15 078?

INSTRUCTION: Round up your answer to the nearest year.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.


STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

  • A=15 078
  • P=3 590
  • i=9,0100
  • n=?

STEP: Transform the simple interest formula for n
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make n the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1i=n

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

n=(15 0783 590)1(9,0100)=35,55555...=36 yearsround UP to the nearest integer

Rounding up to the nearest year, it will take 36 years to reach the goal of saving R15 078.

The answer, rounded up to the nearest integer, is 36 years.


Submit your answer as:

Simple interest investment

David wants to invest R6 740 at a simple interest rate of 12,0% p.a.

How many years will it take for the money to grow to R24 938?

INSTRUCTION: Round up your answer to the nearest year.
Answer: It will take years.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.


STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

  • A=24 938
  • P=6 740
  • i=12,0100
  • n=?

STEP: Transform the simple interest formula for n
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make n the subject of the formula:

A=P(1+in)AP=1+inAP1=inAP1i=n

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

n=(24 9386 740)1(12,0100)=22,5=23 yearsround UP to the nearest integer

Rounding up to the nearest year, it will take 23 years to reach the goal of saving R24 938.

The answer, rounded up to the nearest integer, is 23 years.


Submit your answer as:

Compound interest rate

Qiniso invests R6 320 into an account which pays out a lump sum at the end of 6 years.

If he gets R7 963,20 at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=7 963,20
  • P=6 320,00
  • i=?
  • n=6

STEP: Transform the compound interest formula for i
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make i the subject of the formula:

A=P(1+i)nAP=(1+i)n(AP)1n=1+i(AP)1n1=i

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

i=(7 963,206 320)161=0,039=3,9% per annum

For the situation in this question, the effective interest rate is 3,9% per annum.


Submit your answer as:

Compound interest rate

Akinlabi invests R2 850 into an account which pays out a lump sum at the end of 10 years.

If he gets R3 847,50 at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=3 847,50
  • P=2 850,00
  • i=?
  • n=10

STEP: Transform the compound interest formula for i
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make i the subject of the formula:

A=P(1+i)nAP=(1+i)n(AP)1n=1+i(AP)1n1=i

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

i=(3 847,502 850)1101=0,03=3,0% per annum

For the situation in this question, the effective interest rate is 3,0% per annum.


Submit your answer as:

Compound interest rate

Carl invests R1 860 into an account which pays out a lump sum at the end of 11 years.

If he gets R3 087,60 at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=3 087,60
  • P=1 860,00
  • i=?
  • n=11

STEP: Transform the compound interest formula for i
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make i the subject of the formula:

A=P(1+i)nAP=(1+i)n(AP)1n=1+i(AP)1n1=i

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

i=(3 087,601 860)1111=0,047=4,7% per annum

For the situation in this question, the effective interest rate is 4,7% per annum.


Submit your answer as:

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

  • Chinedu goes to We Pay More Bank and deposits R6 000,00 into an account which pays 11,5% p.a. compounded every month. After 8 years the interest rate changes to 11,5% p.a., now compounded semi-annually. Chinedu leaves the money in the account for a total of 15 years.

  1. What change happened to the account, and in which year did it happen?
    Answer: What changes?
    The change happened after years.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.
    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    There is a lot of information in story about the investment. The key part for this question is: ' After 8 years the interest rate changes... now compounded semi-annually .'

    During this investment, the compounding period changed and it happens 8 years after Chinedu opens the account


    Submit your answer as: and
  2. How many years passed from the change in the account to the final year?
    Answer: years
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The investment lasts a total of 15 years, and the change happens 8 years after Chinedu opens the account. How many years are missing at the end?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The investment lasts a total of 15 years; and from above we know that there was a change 8 years after the account was opened. Then the number of years from the change to the end must be 158=7.

    It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 7 years at the end of the investment period.


    Submit your answer as:
  3. How often was interest compounded during the 3rd year of the investment?
    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    In the story, the bank makes a change to the interest rate. When does this happen? In the 3rd year which interest information applies?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The compounding period for the account changes after 8 years; the 3rd year is before the change occurs. Therefore in the 3rd year is interest is paid once each month.


    Submit your answer as:

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

  • Oladapo goes to Make Some Money Bank and deposits R5 250,00 into an account which pays 9,0% p.a. compounded semi-annually. After 4 years the bank changes the interest rate to 7,5% p.a., still compounded semi-annually. Oladapo leaves the money in the account for a total of 10 years.

  1. What change happened to the account, and in which year did it happen?
    Answer: What changes?
    The change happened after years.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.
    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    There is a lot of information in story about the investment. The key part for this question is: ' After 4 years the bank changes the interest rate .'

    During this investment, the interest percentage rate changed and it happens 4 years after Oladapo opens the account


    Submit your answer as: and
  2. How many years passed from the change in the account to the final year?
    Answer: years
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The investment lasts a total of 10 years, and the change happens 4 years after Oladapo opens the account. How many years are missing at the end?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The investment lasts a total of 10 years; and from above we know that there was a change 4 years after the account was opened. Then the number of years from the change to the end must be 104=6.

    It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 6 years at the end of the investment period.


    Submit your answer as:
  3. What was the interest rate in the eighth year of the investment?
    Answer: %
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    In the story, the bank makes a change to the interest rate. When does this happen? In the eighth year which interest information applies?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The interest rate for the account changes after 4 years; the eighth year is after the change occurs. Therefore the interest rate in the eighth year is the original rate, 7,5%.


    Submit your answer as:

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

  • Save Here Bank offers a savings account which receives 11,8% interest p.a. compounded monthly. Abodunrin decides to open an account, and deposits R8 250,00. His account accrues interest for 7 years until he takes R2 500 from the account. After that, Abodunrin leaves the money in the account until 12 years after his original deposit.

  1. What change happened to the account, and in which year did it happen?
    Answer: What changes?
    The change happened after years.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.
    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    There is a lot of information in story about the investment. The key part for this question is: 'His account accrues interest for 7 years until he takes R2 500 from the account.'

    During this investment, Abodunrin took money out of the account and it happens 7 years after Abodunrin opens the account


    Submit your answer as: and
  2. How many years passed from the change in the account to the final year?
    Answer: years
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The investment lasts a total of 12 years, and the change happens 7 years after Abodunrin opens the account. How many years are missing at the end?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The investment lasts a total of 12 years; and from above we know that there was a change 7 years after the account was opened. Then the number of years from the change to the end must be 127=5.

    It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 5 years at the end of the investment period.


    Submit your answer as:
  3. During this investment, how much money did Abodunrin deposit in total?
    Answer: R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    In the beginning, Abodunrin deposits R8 250,00. How is this amount changed when he takes from the account?
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    Abodunrin opens the account with a deposit of R8 250,00. Later he takes R2 500 out of the account. Therefore, the total amount of money he put into the account is 8 2502 500=5 750.


    Submit your answer as:

Finding the length of an investment

Some years ago, a banker opened a savings account at North Gauteng Bank. The principal amount was R3 200 but now the account has a value of R4 149,26. For how many years was the money in the account if the account pays 6,71% compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.
Answer: The money was in the account for years.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.


STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

A=P(1+i)nWhere: A=4 149,26P=3 200i=6,71%=0,0671n=?

STEP: Substitute the values into the formula and begin solving for n
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of n. Substitute in the known values and make n the subject of the formula:

4 149,26=3 200(1+0,0671)nthis equation by 3 200divide both side of4 149,26=3 200(1,0671)n4 149,263 200=(1,0671)n

STEP: Rewrite the equation in logarithmic form and calculate n
[−1 point ⇒ 0 / 3 points left]

To make n the subject of the formula, we convert the equation from exponential form to logarithmic form:

n=log1,0671(4 149,263 200)n=log(1,2966...)log1,0671

Use a calculator to evaluate the logarithm:

n=4

Therefore, the banker left the money in the account for 4 years.


Submit your answer as:

Finding the length of an investment

Some years ago, a woman opened a savings account at KZN South Bank. The principal amount was R3 000 but now the account has a value of R4 681,83. For how many years was the money in the account if the account pays 7,7% compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.
Answer: The money was in the account for years.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.


STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

A=P(1+i)nWhere: A=4 681,83P=3 000i=7,7%=0,077n=?

STEP: Substitute the values into the formula and begin solving for n
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of n. Substitute in the known values and make n the subject of the formula:

4 681,83=3 000(1+0,077)nthis equation by 3 000divide both side of4 681,83=3 000(1,077)n4 681,833 000=(1,077)n

STEP: Rewrite the equation in logarithmic form and calculate n
[−1 point ⇒ 0 / 3 points left]

To make n the subject of the formula, we convert the equation from exponential form to logarithmic form:

n=log1,077(4 681,833 000)n=log(1,5606...)log1,077

Use a calculator to evaluate the logarithm:

n=6

Therefore, the woman left the money in the account for 6 years.


Submit your answer as:

Finding the length of an investment

Some years ago, a woman opened a savings account at Limpopo West Bank. The principal amount was R2 100 but now the account has a value of R4 419,85. For how many years was the money in the account if the account pays 8,62% compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.
Answer: The money was in the account for years.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.


STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

A=P(1+i)nWhere: A=4 419,85P=2 100i=8,62%=0,0862n=?

STEP: Substitute the values into the formula and begin solving for n
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of n. Substitute in the known values and make n the subject of the formula:

4 419,85=2 100(1+0,0862)nthis equation by 2 100divide both side of4 419,85=2 100(1,0862)n4 419,852 100=(1,0862)n

STEP: Rewrite the equation in logarithmic form and calculate n
[−1 point ⇒ 0 / 3 points left]

To make n the subject of the formula, we convert the equation from exponential form to logarithmic form:

n=log1,0862(4 419,852 100)n=log(2,1046...)log1,0862

Use a calculator to evaluate the logarithm:

n=9

Therefore, the woman left the money in the account for 9 years.


Submit your answer as:

Compound interest investment

Amaka wants to invest some money at a compound interest rate of 10,1% p.a.

How much money should be invested if she wants to reach a sum of R36 000 in 4 years' time?

INSTRUCTION: Round up your answer to the nearest rand.
Answer: Amaka must invest R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=R36 000
  • P=?
  • i=10,1%=10,1100
  • n=4

STEP: Transform the equation for P
[−1 point ⇒ 1 / 3 points left]

The question states that the investment grows at a compound interest rate. So we will use the compound interest formula. To determine the amount she must invest, we can to make P the subject of the formula:

A=P(1+i)nA(1+i)n=P

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]
36 000(1+10,1100)4=P36 000(1,101)4=PP=R24 499,27

The result of the calculation shows that Amaka must invest R24 499,27 if she is going to reach her goal of R36 000 in 4 years. The question tells us to "round up your answer to the nearest rand," so the final answer is R24 500.

NOTE:

Rounding up is important here. In this question, Amaka is aiming to reach a final goal of R36 000. If we round the answer down, then after 4 years the total amount of money will not quite reach R36 000 - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is R24 500.


Submit your answer as:

Compound interest investment

Mathe wants to invest some money at a compound interest rate of 6,1% p.a.

How much money should be invested if she wants to reach a sum of R26 800 in 6 years' time?

INSTRUCTION: Round up your answer to the nearest rand.
Answer: Mathe must invest R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=R26 800
  • P=?
  • i=6,1%=6,1100
  • n=6

STEP: Transform the equation for P
[−1 point ⇒ 1 / 3 points left]

The question states that the investment grows at a compound interest rate. So we will use the compound interest formula. To determine the amount she must invest, we can to make P the subject of the formula:

A=P(1+i)nA(1+i)n=P

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]
26 800(1+6,1100)6=P26 800(1,061)6=PP=R18 786,35

The result of the calculation shows that Mathe must invest R18 786,35 if she is going to reach her goal of R26 800 in 6 years. The question tells us to "round up your answer to the nearest rand," so the final answer is R18 787.

NOTE:

Rounding up is important here. In this question, Mathe is aiming to reach a final goal of R26 800. If we round the answer down, then after 6 years the total amount of money will not quite reach R26 800 - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is R18 787.


Submit your answer as:

Compound interest investment

Nnenna wants to invest some money at a compound interest rate of 8,8% p.a.

How much money should be invested if she wants to reach a sum of R11 800 in 6 years' time?

INSTRUCTION: Round up your answer to the nearest rand.
Answer: Nnenna must invest R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

  • A=R11 800
  • P=?
  • i=8,8%=8,8100
  • n=6

STEP: Transform the equation for P
[−1 point ⇒ 1 / 3 points left]

The question states that the investment grows at a compound interest rate. So we will use the compound interest formula. To determine the amount she must invest, we can to make P the subject of the formula:

A=P(1+i)nA(1+i)n=P

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]
11 800(1+8,8100)6=P11 800(1,088)6=PP=R7 113,91

The result of the calculation shows that Nnenna must invest R7 113,91 if she is going to reach her goal of R11 800 in 6 years. The question tells us to "round up your answer to the nearest rand," so the final answer is R7 114.

NOTE:

Rounding up is important here. In this question, Nnenna is aiming to reach a final goal of R11 800. If we round the answer down, then after 6 years the total amount of money will not quite reach R11 800 - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is R7 114.


Submit your answer as:

Timeline questions: changing interest rates

Nnenne goes to Save Here Bank and deposits R8 750,00 into an account which pays 11,8% p.a. compounded every month. After 6 years the bank changes the interest rate to 12,7% p.a., still compounded every month.

  1. How much money will Nnenne have in her account 11 years after the original investment?

    Answer: Total amount = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.


    STEP: <no title>
    [−3 points ⇒ 3 / 6 points left]

    This question describes a 11 year period which is broken into two sections because of the change in the interest rate: the first section lasts 6 years (while the interest rate is 11,8%); the second section lasts 5 years (when the rate is 12,7%).

    Determine what has been provided and what is required for the first 6 years:

    P=8 750i=11,8%=0,118n=6A= ?

    This is a compound interest problem:

    A=P(1+i)n

    Solve the first part of the problem. Remember that we must multiply the number of years by 12 and divide the interest rate by 12 because the interest is compounded every month.

    A=8 750,0(1+0,11812)(6×12)A=8 750,0(1,009...)72=8 750,0(2,022...)=R17 700,54

    This is the amount after 6 years.


    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 5 years. All of the money from the above calculation remains in the account - it is the starting point for the last 5 years.

    P=R17 700,54i=12,7%=0,127n=5A= ?

    Now solve the second part of the problem, the last 5 years (again, we must adjust the interest rate and the the number of years by 12.):

    A=17 700,54(1+0,12712)(5×12)=17 700,54(1,010...)60=17 700,54(1,880...)=R33 290,05

    After 11 years, Nnenne will have R33 290,05 in her account.


    Submit your answer as:
  2. What is the total amount of interest the bank pays into her account?

    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The amount of interest is what the bank paid into the account on top of what Nnenne put into the account at the beginning.


    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    Use the answer from the work above to calculate how much interest the bank paid.

    Interest earned =R33 290,05R8 750,00=R24 540,05

    The amount of interest the bank pays is R24 540,05.


    Submit your answer as:

Timeline questions: changing interest rates

Anthea goes to We Pay More Bank and deposits R6 750,00 into an account which pays 11,1% p.a. compounded quarterly. After 6 years the bank changes the interest rate to 9,9% p.a., still compounded quarterly.

  1. How much money will Anthea have in her account 10 years after the original investment?

    Answer: Total amount = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.


    STEP: <no title>
    [−3 points ⇒ 3 / 6 points left]

    This question describes a 10 year period which is broken into two sections because of the change in the interest rate: the first section lasts 6 years (while the interest rate is 11,1%); the second section lasts 4 years (when the rate is 9,9%).

    Determine what has been provided and what is required for the first 6 years:

    P=6 750i=11,1%=0,111n=6A= ?

    This is a compound interest problem:

    A=P(1+i)n

    Solve the first part of the problem. Remember that we must multiply the number of years by 4 and divide the interest rate by 4 because the interest is compounded quarterly.

    A=6 750,0(1+0,1114)(6×4)A=6 750,0(1,027...)24=6 750,0(1,928...)=R13 019,77

    This is the amount after 6 years.


    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 4 years. All of the money from the above calculation remains in the account - it is the starting point for the last 4 years.

    P=R13 019,77i=9,9%=0,099n=4A= ?

    Now solve the second part of the problem, the last 4 years (again, we must adjust the interest rate and the the number of years by 4.):

    A=13 019,77(1+0,0994)(4×4)=13 019,77(1,024...)16=13 019,77(1,478...)=R19 252,63

    After 10 years, Anthea will have R19 252,63 in her account.


    Submit your answer as:
  2. What is the total amount of interest the bank pays into her account?

    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The amount of interest is what the bank paid into the account on top of what Anthea put into the account at the beginning.


    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    Use the answer from the work above to calculate how much interest the bank paid.

    Interest earned =R19 252,63R6 750,00=R12 502,63

    The amount of interest the bank pays is R12 502,63.


    Submit your answer as:

Timeline questions: changing interest rates

Bianca goes to Make Some Money Bank and deposits R4 250,00 into an account which pays 7,7% p.a. compounded two times per year. After 5 years the bank changes the interest rate to 9,2% p.a., still compounded two times per year.

  1. How much money will Bianca have in her account 8 years after the original investment?

    Answer: Total amount = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.


    STEP: <no title>
    [−3 points ⇒ 3 / 6 points left]

    This question describes a 8 year period which is broken into two sections because of the change in the interest rate: the first section lasts 5 years (while the interest rate is 7,7%); the second section lasts 3 years (when the rate is 9,2%).

    Determine what has been provided and what is required for the first 5 years:

    P=4 250i=7,7%=0,077n=5A= ?

    This is a compound interest problem:

    A=P(1+i)n

    Solve the first part of the problem. Remember that we must multiply the number of years by 2 and divide the interest rate by 2 because the interest is compounded two times per year.

    A=4 250,0(1+0,0772)(5×2)A=4 250,0(1,038...)10=4 250,0(1,459...)=R6 200,89

    This is the amount after 5 years.


    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Determine what has been provided and what is required for the next 3 years. All of the money from the above calculation remains in the account - it is the starting point for the last 3 years.

    P=R6 200,89i=9,2%=0,092n=3A= ?

    Now solve the second part of the problem, the last 3 years (again, we must adjust the interest rate and the the number of years by 2.):

    A=6 200,89(1+0,0922)(3×2)=6 200,89(1,046...)6=6 200,89(1,309...)=R8 121,65

    After 8 years, Bianca will have R8 121,65 in her account.


    Submit your answer as:
  2. What is the total amount of interest the bank pays into her account?

    Answer: Interest earned = R
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The amount of interest is what the bank paid into the account on top of what Bianca put into the account at the beginning.


    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    Use the answer from the work above to calculate how much interest the bank paid.

    Interest earned =R8 121,65R4 250,00=R3 871,65

    The amount of interest the bank pays is R3 871,65.


    Submit your answer as:

Compound interest investment

An amount of R2 160 is invested in a savings account which pays a compound interest rate of 8,8% p.a.

Calculate the balance accumulated by the end of 6 years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.


STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=2 160
  • i=8,8100
  • n=6

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

A=P(1+i)n=2 160(1+8,8100)6=2 160(1,088)6=R3 582,84

If you have an answer close to this, it is likely that you rounded off during the calculation. However, you should not round off until you reach the end of the calculation - let your calculator hold all of the decimal digits that come as you do the calculation until the end.

The balance after years is R3 582,84.


Submit your answer as:

Compound interest investment

An amount of R4 370 is invested in a savings account which pays a compound interest rate of 8,3% p.a.

Calculate the balance accumulated by the end of 3 years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.


STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=4 370
  • i=8,3100
  • n=3

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

A=P(1+i)n=4 370(1+8,3100)3=4 370(1,083)3=R5 550,94

If you have an answer close to this, it is likely that you rounded off during the calculation. However, you should not round off until you reach the end of the calculation - let your calculator hold all of the decimal digits that come as you do the calculation until the end.

The balance after years is R5 550,94.


Submit your answer as:

Compound interest investment

An amount of R4 190 is invested in a savings account which pays a compound interest rate of 6,8% p.a.

Calculate the balance accumulated by the end of 7 years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.


STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

  • A=?
  • P=4 190
  • i=6,8100
  • n=7

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

A=P(1+i)n=4 190(1+6,8100)7=4 190(1,068)7=R6 640,68

If you have an answer close to this, it is likely that you rounded off during the calculation. However, you should not round off until you reach the end of the calculation - let your calculator hold all of the decimal digits that come as you do the calculation until the end.

The balance after years is R6 640,68.


Submit your answer as:

Timelines: population growth

Farmer Mcebisi starts working with a beehive in his orchard. The beehive contains 52 000 bees. The number of bees in the hive grows by 2,65% each year. After 7 years farmer Mcebisi gives 9 000 of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate (2,65% every year).

  1. How many bees will be in farmer Mcebisi's hive 4 years after he gave some of his bees away?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in Mcebisi's hive (rounded to the nearest whole number).

    one-of
    type(numeric.abserror(0.5))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.


    STEP: Calculate the number of bees after 7 years
    [−2 points ⇒ 2 / 4 points left]

    Populations grow like compound interest, so we should use this formula: A=P(1+i)n.

    In this case we will use the equation as follows:

    • P is the number of bees at the beginning =52 000
    • i is the rate of growth of th hive =2,65%=0,0265
    • n is the number of years =7
    • A=?

    Solve the first part of the problem:

    A=52 000(1+0,0265)7=62 447,638...62 448
    TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So, there are approximately 62 448 bees in the hive after 7 years.


    STEP: Calculate the number of bees after 11 years
    [−2 points ⇒ 0 / 4 points left]

    At this point the farmer takes away (subtracts) 9 000 bees from his hive. You must subtract now because Mcebisi is taking bees out of the hive.

    P=62 4489 000=53 448i=2,65%=0,0265n=4A= ?

    Now calculate how many bees there will be after the last 4 years.

    A=53 448(1+0,0265)4=59 342,696...59 343

    So, after 11 years, farmer Mcebisi will have 59 343 bees in his beehive.

    NOTE:

    You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

    A=52 000(1+0,0265)119 000(1+0,0265)4=59 342,294...59 342

    This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.


    Submit your answer as:
  2. There were 9 000 in the second hive when it started. How many bees will be in the second hive after 4 years?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in the second hive (rounded to the nearest whole number).

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).


    STEP: Calculate the number of bees in the second hive after 4 years
    [−2 points ⇒ 0 / 2 points left]

    For this question, remember that the second hive starts with 9 000 bees, but it only starts 7 years after the first hive. Therefore the second hive only grows for 4 years.

    P=9 000i=2,65%=0,0265n=4A= ?
    A=9 000(1+0,0265)4=9 992,595...9 993

    So, after 4 years, the second hive had approximately 9 993 bees in it.


    Submit your answer as:
  3. A single bee can make about 0,56 grams of honey in a summer. If 90% of the bees in the colony produce honey, how much honey did the two hives produce in the eleventh year after farmer Mcebisi started working with bees?

    INSTRUCTION: Write your answer in kilograms and then round it to two decimal places.
    Answer:

    The amount of honey produced is kg.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    How many bees are there in both hives after 11 years?


    STEP: Calculate the number of bees that produce honey
    [−1 point ⇒ 1 / 2 points left]

    90% of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate 90% (0,90) of the total number of bees:

    0,90×(59 343+9 993)=62 402,462 402
    TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So there are about 62 402 bees making honey.


    STEP: Calculate how much honey these bees make, by multiplying
    [−0 points ⇒ 1 / 2 points left]

    Each of the 62 402 bees produces 0,56 grams of honey. To find the total amount of honey, we just multiply:

    0,56×62 402=34 945,12

    STEP: Convert grams to kilograms
    [−1 point ⇒ 0 / 2 points left]

    But we aren't done yet because the question asks us to write the answer in kilograms. Every kg is equal to 1 000 g.

    The amount of honey (in kg) is:

    34 945,121000=34,9451234,95 kg

    The two hives produce approximately 34,95 kg of honey.


    Submit your answer as:

Timelines: population growth

Farmer Anathi starts working with a beehive in his orchard. The beehive contains 59 000 bees. The number of bees in the hive grows by 2,65% each year. After 7 years farmer Anathi gives 10 500 of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate (2,65% every year).

  1. How many bees will be in farmer Anathi's hive 5 years after he gave some of his bees away?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in Anathi's hive (rounded to the nearest whole number).

    one-of
    type(numeric.abserror(0.5))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.


    STEP: Calculate the number of bees after 7 years
    [−2 points ⇒ 2 / 4 points left]

    Populations grow like compound interest, so we should use this formula: A=P(1+i)n.

    In this case we will use the equation as follows:

    • P is the number of bees at the beginning =59 000
    • i is the rate of growth of th hive =2,65%=0,0265
    • n is the number of years =7
    • A=?

    Solve the first part of the problem:

    A=59 000(1+0,0265)7=70 854,051...70 854
    TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So, there are approximately 70 854 bees in the hive after 7 years.


    STEP: Calculate the number of bees after 12 years
    [−2 points ⇒ 0 / 4 points left]

    At this point the farmer takes away (subtracts) 10 500 bees from his hive. You must subtract now because Anathi is taking bees out of the hive.

    P=70 85410 500=60 354i=2,65%=0,0265n=5A= ?

    Now calculate how many bees there will be after the last 5 years.

    A=60 354(1+0,0265)5=68 786,122...68 786

    So, after 12 years, farmer Anathi will have 68 786 bees in his beehive.

    NOTE:

    You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

    A=59 000(1+0,0265)1210 500(1+0,0265)5=68 786,180...68 786

    This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.


    Submit your answer as:
  2. There were 10 500 in the second hive when it started. How many bees will be in the second hive after 5 years?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in the second hive (rounded to the nearest whole number).

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).


    STEP: Calculate the number of bees in the second hive after 5 years
    [−2 points ⇒ 0 / 2 points left]

    For this question, remember that the second hive starts with 10 500 bees, but it only starts 7 years after the first hive. Therefore the second hive only grows for 5 years.

    P=10 500i=2,65%=0,0265n=5A= ?
    A=10 500(1+0,0265)5=11 966,966...11 967

    So, after 5 years, the second hive had approximately 11 967 bees in it.


    Submit your answer as:
  3. A single bee can make about 0,56 grams of honey in a summer. If 90% of the bees in the colony produce honey, how much honey did the two hives produce in the 12th year after farmer Anathi started working with bees?

    INSTRUCTION: Write your answer in kilograms and then round it to two decimal places.
    Answer:

    The amount of honey produced is kg.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    How many bees are there in both hives after 12 years?


    STEP: Calculate the number of bees that produce honey
    [−1 point ⇒ 1 / 2 points left]

    90% of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate 90% (0,90) of the total number of bees:

    0,90×(68 786+11 967)=72 677,772 678
    TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So there are about 72 678 bees making honey.


    STEP: Calculate how much honey these bees make, by multiplying
    [−0 points ⇒ 1 / 2 points left]

    Each of the 72 678 bees produces 0,56 grams of honey. To find the total amount of honey, we just multiply:

    0,56×72 678=40 699,68

    STEP: Convert grams to kilograms
    [−1 point ⇒ 0 / 2 points left]

    But we aren't done yet because the question asks us to write the answer in kilograms. Every kg is equal to 1 000 g.

    The amount of honey (in kg) is:

    40 699,681000=40,6996840,70 kg

    The two hives produce approximately 40,70 kg of honey.


    Submit your answer as:

Timelines: population growth

Farmer Lulamile starts working with a beehive in his orchard. The beehive contains 58 000 bees. The number of bees in the hive grows by 2,32% each year. After 7 years farmer Lulamile gives 9 500 of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate (2,32% every year).

  1. How many bees will be in farmer Lulamile's hive 5 years after he gave some of his bees away?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in Lulamile's hive (rounded to the nearest whole number).

    one-of
    type(numeric.abserror(0.5))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.


    STEP: Calculate the number of bees after 7 years
    [−2 points ⇒ 2 / 4 points left]

    Populations grow like compound interest, so we should use this formula: A=P(1+i)n.

    In this case we will use the equation as follows:

    • P is the number of bees at the beginning =58 000
    • i is the rate of growth of th hive =2,32%=0,0232
    • n is the number of years =7
    • A=?

    Solve the first part of the problem:

    A=58 000(1+0,0232)7=68 100,721...68 101
    TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So, there are approximately 68 101 bees in the hive after 7 years.


    STEP: Calculate the number of bees after 12 years
    [−2 points ⇒ 0 / 4 points left]

    At this point the farmer takes away (subtracts) 9 500 bees from his hive. You must subtract now because Lulamile is taking bees out of the hive.

    P=68 1019 500=58 601i=2,32%=0,0232n=5A= ?

    Now calculate how many bees there will be after the last 5 years.

    A=58 601(1+0,0232)5=65 721,532...65 722

    So, after 12 years, farmer Lulamile will have 65 722 bees in his beehive.

    NOTE:

    You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

    A=58 000(1+0,0232)129 500(1+0,0232)5=65 721,220...65 721

    This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.


    Submit your answer as:
  2. There were 9 500 in the second hive when it started. How many bees will be in the second hive after 5 years?

    INSTRUCTION: Round your answer to the nearest whole number.
    Answer:

    There will be bees in the second hive (rounded to the nearest whole number).

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).


    STEP: Calculate the number of bees in the second hive after 5 years
    [−2 points ⇒ 0 / 2 points left]

    For this question, remember that the second hive starts with 9 500 bees, but it only starts 7 years after the first hive. Therefore the second hive only grows for 5 years.

    P=9 500i=2,32%=0,0232n=5A= ?
    A=9 500(1+0,0232)5=10 654,332...10 654

    So, after 5 years, the second hive had approximately 10 654 bees in it.


    Submit your answer as:
  3. A single bee can make about 0,56 grams of honey in a summer. If 90% of the bees in the colony produce honey, how much honey did the two hives produce in the 12th year after farmer Lulamile started working with bees?

    INSTRUCTION: Write your answer in kilograms and then round it to two decimal places.
    Answer:

    The amount of honey produced is kg.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    How many bees are there in both hives after 12 years?


    STEP: Calculate the number of bees that produce honey
    [−1 point ⇒ 1 / 2 points left]

    90% of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate 90% (0,90) of the total number of bees:

    0,90×(65 722+10 654)=68 738,468 738
    TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

    So there are about 68 738 bees making honey.


    STEP: Calculate how much honey these bees make, by multiplying
    [−0 points ⇒ 1 / 2 points left]

    Each of the 68 738 bees produces 0,56 grams of honey. To find the total amount of honey, we just multiply:

    0,56×68 738=38 493,28

    STEP: Convert grams to kilograms
    [−1 point ⇒ 0 / 2 points left]

    But we aren't done yet because the question asks us to write the answer in kilograms. Every kg is equal to 1 000 g.

    The amount of honey (in kg) is:

    38 493,281000=38,4932838,49 kg

    The two hives produce approximately 38,49 kg of honey.


    Submit your answer as:

2. Depreciation

Increasing values: finding i

After 7 years an investment doubles in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Read the question carefully and write down the given information.
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 2 times as much as at the beginning:

  • A=2×P
  • P=P
  • n=7
  • i=?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

A=P(1+i)n

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

2P=P(1+i100)72=(1+i100)7each other!The P's cancel

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for i. The first step is to use what you know about exponents - how can you cancel the power of 7 in the equation? You must use the seventh root, as shown below:

(2)7=(1+i100)771,1040=1+i1000,1040=i100(0,1040)×100=ii=10,4089


STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

Therefore i=10,41%

Submit your answer as:

Simple depreciation: when an asset goes to zero value!

A small business buys a photocopier for R11 000. For the tax return, the owner depreciates this asset over 7 years on a straight-line basis. Each year the company's owner must fill in the value of the photocopier on this tax form. What amount will he put on his tax form after 4 year(s)?

Answer:R
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
In this question, the value of the photocopier decreases each and every year until the value has become R0 after 7 years. Start by calculating how much the value decreases for each year.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the photocopier to depreciate to R0 after 7 years. Since he is using straight-line depreciation, the photocopier will decrease by the same amount every year. This must be the value of the photocopier divided by the number of years: 11 000÷7= R1 571,43 per year.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the photocopier lost in 4 year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

value lost=4(1 571,43)=6 285,72


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the photocopier after 4 year/s is:

11 0006 285,72=R4 714,28


Submit your answer as:

Compound interest: finding i

Nthabiseng deposits R3 500 into a savings account which earns compound interest. The account grows to a value of R4 467,22 in 4 years. What is the interest rate for this account?

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

  • A=R4 467,22
  • P=R3 500
  • n=4
  • i=?

The question says, "earns compound interest," so we need to use the compound interest formula. Substitute in what you know!

A=P(1+i)n4 467,22=3 500(1+i)4

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for i. The first step is to divide by 3 500. Then we will deal with the exponent.

4 467,22=3 500(1+i)44 467,223 500=(1+i)41,27634=(1+i)4

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

At this point we must use our exponent skills from earlier in the year! To remove the exponent on the right side of the equation, we need to take a root to kill the exponent of 4. (Rather than a root, we can also use an exponent of 14 - it means the same thing.)

1,276344=(1+i)441,06289=1+i0,06289=i

Finally, change the decimal into a percentage (multiply by 100) and then round to two decimal places: i=100(0,06289)=6,29%.


Submit your answer as:

Compound depreciation: population loss

The number of cormorants at the Berg river mouth is decreasing at a compound rate of 15% p.a. If there are now 6 000 cormorants, how many will there be in 19 years' time?

Answer: There will be cormorants.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

P=6 000i=15%=0,15n=19A is required

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

A=P(1i)nA=6 000(10,15)19

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=6 000(0,85)19=6 000×0,045599...off yet!Don't round=273,59

There will be approximately 274 cormorants in 19 years' time. (You must round to the nearest integer because you cannot have 273,59 cormorants!)


Submit your answer as:

Simple depreciation

A boat is worth R64 000 now. It depreciates at a rate of 14% p.a. following the straight-line depreciation method. What is the boat worth in 5 years' time?

Answer: The boat will be worth R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.


STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

the value now=64 000rate of depreciation=14%=0,14number of years=5the final value=?

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

A=P(1in)A=64 000(1(0,14×5))

STEP: Solve for A
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=64 000(10,7)=64 000(0,3)=19 200

In 5 years' time the boat will be worth R19 200.


Submit your answer as:

Compound depreciation

Farmer Akinbode buys a tractor for R98 000. The tractor depreciates by 21% per year where the depreciation happens at a compound rate. What is the depreciated value of the tractor after 14 years?

Answer:

The depreciated value of the tractor is R .

numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

original value=R98 000rate of depreciation=21%=0,21number of years=14the final value=?

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

A=P(1i)nA=98 000(10,21)14

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=98 000(0,79)14=98 000×0,03687...=3 614,1431...

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the tractor after 14 years is R3 614,14.


Submit your answer as:

Simple interest: finding i

Hein deposits R5 000 into a savings account which earns simple interest. The account grows to a value of R7 100,00 in 7 years. Determine the value of i, the rate of interest paid by the bank.

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

  • A=7100
  • P=5 000
  • n=7
  • i=?

The question says, "earns simple interest," so we need to use the simple interest formula. Substitute in what you know!

A=P(1+in)7100=5 000(1+i(7))

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for i. The first step is to divide by 5 000 which will let us get into the brackets. Then solve it!

7100=5 000(1+7i)71005 000=(1+7i)1,42=1+7i0,42=7i0,427=i0,06=i

Finally, multiply by 100 to change into a percentage. Then round to one decimal place if necessary. i=100(0,06)=6,0%.


Submit your answer as:

Exercises

Increasing values: finding i

After 5 years an investment doubles in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Read the question carefully and write down the given information.
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 2 times as much as at the beginning:

  • A=2×P
  • P=P
  • n=5
  • i=?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

A=P(1+i)n

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

2P=P(1+i100)52=(1+i100)5each other!The P's cancel

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for i. The first step is to use what you know about exponents - how can you cancel the power of 5 in the equation? You must use the fifth root, as shown below:

(2)5=(1+i100)551,1486=1+i1000,1486=i100(0,1486)×100=ii=14,8698


STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

Therefore i=14,87%

Submit your answer as:

Increasing values: finding i

After 4 years an investment doubles in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Read the question carefully and write down the given information.
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 2 times as much as at the beginning:

  • A=2×P
  • P=P
  • n=4
  • i=?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

A=P(1+i)n

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

2P=P(1+i100)42=(1+i100)4each other!The P's cancel

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for i. The first step is to use what you know about exponents - how can you cancel the power of 4 in the equation? You must use the fourth root, as shown below:

(2)4=(1+i100)441,1892=1+i1000,1892=i100(0,1892)×100=ii=18,9207


STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

Therefore i=18,92%

Submit your answer as:

Increasing values: finding i

After 9 years an investment quadruples (gets 4 times bigger) in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Read the question carefully and write down the given information.
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 4 times as much as at the beginning:

  • A=4×P
  • P=P
  • n=9
  • i=?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

A=P(1+i)n

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

4P=P(1+i100)94=(1+i100)9each other!The P's cancel

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for i. The first step is to use what you know about exponents - how can you cancel the power of 9 in the equation? You must use the ninth root, as shown below:

(4)9=(1+i100)991,1665=1+i1000,1665=i100(0,1665)×100=ii=16,6529


STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

Therefore i=16,65%

Submit your answer as:

Simple depreciation: when an asset goes to zero value!

A small business buys a photocopier for R12 000. For the tax return, the owner depreciates this asset over 4 years following the straight-line method. Each year the company's owner must fill in the value of the photocopier on this tax form. What amount will he put on his tax form after 2 year(s)?

Answer:R
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
In this question, the value of the photocopier decreases each and every year until the value has become R0 after 4 years. Start by calculating how much the value decreases for each year.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the photocopier to depreciate to R0 after 4 years. Since he is using straight-line depreciation, the photocopier will decrease by the same amount every year. This must be the value of the photocopier divided by the number of years: 12 000÷4= R3 000 per year.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the photocopier lost in 2 year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

value lost=2(3 000)=6 000


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the photocopier after 2 year/s is:

12 0006 000=R6 000


Submit your answer as:

Simple depreciation: when an asset goes to zero value!

A small business buys a projector for R10 000. For the tax return, the owner depreciates this asset over 10 years using a straight-line depreciation method. Each year the company's owner must fill in the value of the projector on this tax form. What amount will he put on his tax form after 6 year(s)?

Answer:R
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
In this question, the value of the projector decreases each and every year until the value has become R0 after 10 years. Start by calculating how much the value decreases for each year.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the photocopier to depreciate to R0 after 10 years. Since he is using straight-line depreciation, the projector will decrease by the same amount every year. This must be the value of the projector divided by the number of years: 10 000÷10= R1 000 per year.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the projector lost in 6 year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

value lost=6(1 000)=6 000


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the projector after 6 year/s is:

10 0006 000=R4 000


Submit your answer as:

Simple depreciation: when an asset goes to zero value!

A small business buys a computer for R3 000. For the tax return, the owner depreciates this asset over 8 years following the straight-line method. Each year the company's owner must fill in the value of the computer on this tax form. What amount will he put on his tax form after 5 year(s)?

Answer:R
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
In this question, the value of the computer decreases each and every year until the value has become R0 after 8 years. Start by calculating how much the value decreases for each year.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the photocopier to depreciate to R0 after 8 years. Since he is using straight-line depreciation, the computer will decrease by the same amount every year. This must be the value of the computer divided by the number of years: 3 000÷8= R375 per year.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the computer lost in 5 year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

value lost=5(375)=1 875


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the computer after 5 year/s is:

3 0001 875=R1 125


Submit your answer as:

Compound interest: finding i

Nthabiseng deposits R4 000 into a savings account which earns compound interest. The account grows to a value of R8 226,40 in 7 years. What is the interest rate for this account?

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

  • A=R8 226,40
  • P=R4 000
  • n=7
  • i=?

The question says, "earns compound interest," so we need to use the compound interest formula. Substitute in what you know!

A=P(1+i)n8 226,4=4 000(1+i)7

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for i. The first step is to divide by 4 000. Then we will deal with the exponent.

8 226,4=4 000(1+i)78 226,44 000=(1+i)72,0566=(1+i)7

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

At this point we must use our exponent skills from earlier in the year! To remove the exponent on the right side of the equation, we need to take a root to kill the exponent of 7. (Rather than a root, we can also use an exponent of 17 - it means the same thing.)

2,05667=(1+i)771,10849=1+i0,10849=i

Finally, change the decimal into a percentage (multiply by 100) and then round to two decimal places: i=100(0,10849)=10,85%.


Submit your answer as:

Compound interest: finding i

Aisha deposits R5 500 into a savings account which gets compound interest. The account grows to a value of R8 612,08 in 6 years. Determine the interest rate the account earns.

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

  • A=R8 612,08
  • P=R5 500
  • n=6
  • i=?

The question says, "gets compound interest," so we need to use the compound interest formula. Substitute in what you know!

A=P(1+i)n8 612,08=5 500(1+i)6

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for i. The first step is to divide by 5 500. Then we will deal with the exponent.

8 612,08=5 500(1+i)68 612,085 500=(1+i)61,56583=(1+i)6

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

At this point we must use our exponent skills from earlier in the year! To remove the exponent on the right side of the equation, we need to take a root to kill the exponent of 6. (Rather than a root, we can also use an exponent of 16 - it means the same thing.)

1,565836=(1+i)661,07759=1+i0,07759=i

Finally, change the decimal into a percentage (multiply by 100) and then round to two decimal places: i=100(0,07759)=7,76%.


Submit your answer as:

Compound interest: finding i

Ros deposits R4 500 into a savings account which earns compound interest. The account grows to a value of R10 082,49 in 8 years. Determine the value of i, the rate of interest paid by the bank.

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

  • A=R10 082,49
  • P=R4 500
  • n=8
  • i=?

The question says, "earns compound interest," so we need to use the compound interest formula. Substitute in what you know!

A=P(1+i)n10 082,49=4 500(1+i)8

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for i. The first step is to divide by 4 500. Then we will deal with the exponent.

10 082,49=4 500(1+i)810 082,494 500=(1+i)82,24055=(1+i)8

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

At this point we must use our exponent skills from earlier in the year! To remove the exponent on the right side of the equation, we need to take a root to kill the exponent of 8. (Rather than a root, we can also use an exponent of 18 - it means the same thing.)

2,240558=(1+i)881,10610=1+i0,10610=i

Finally, change the decimal into a percentage (multiply by 100) and then round to two decimal places: i=100(0,10610)=10,61%.


Submit your answer as:

Compound depreciation: population loss

The number of flamingos at the Bloukrans river mouth is decreasing at a compound rate of 6% p.a. If there are now 5 000 flamingos, how many will there be in 6 years' time?

Answer: There will be flamingos.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

P=5 000i=6%=0,06n=6A is required

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

A=P(1i)nA=5 000(10,06)6

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=5 000(0,94)6=5 000×0,68987...off yet!Don't round=3 449,35

There will be approximately 3 449 flamingos in 6 years' time. (You must round to the nearest integer because you cannot have 3 449,35 flamingos!)


Submit your answer as:

Compound depreciation: population loss

The number of pelicans at the Amanzimtoti river mouth is decreasing at a compound rate of 9% p.a. If there are now 6 000 pelicans, how many will there be in 18 years' time?

Answer: There will be pelicans.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

P=6 000i=9%=0,09n=18A is required

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

A=P(1i)nA=6 000(10,09)18

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=6 000(0,91)18=6 000×0,183124...off yet!Don't round=1 098,74

There will be approximately 1 099 pelicans in 18 years' time. (You must round to the nearest integer because you cannot have 1 098,74 pelicans!)


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Compound depreciation: population loss

The number of flamingos at the Bloukrans river mouth is decreasing at a compound rate of 5% p.a. If there are now 15 000 flamingos, how many will there be in 10 years' time?

Answer: There will be flamingos.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

P=15 000i=5%=0,05n=10A is required

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

A=P(1i)nA=15 000(10,05)10

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=15 000(0,95)10=15 000×0,598737...off yet!Don't round=8 981,06

There will be approximately 8 981 flamingos in 10 years' time. (You must round to the nearest integer because you cannot have 8 981,06 flamingos!)


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Simple depreciation

A car is worth R73 000 now. It depreciates at a rate of 9% p.a. on a straight-line basis. What is the car worth in 9 years' time?

Answer: The car will be worth R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.


STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

the value now=73 000rate of depreciation=9%=0,09number of years=9the final value=?

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

A=P(1in)A=73 000(1(0,09×9))

STEP: Solve for A
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=73 000(10,81)=73 000(0,19)=13 870

In 9 years' time the car will be worth R13 870.


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Simple depreciation

A delivery van is worth R94 000 now. It depreciates at a rate of 6% p.a. following the straight-line depreciation method. What is the delivery van worth in 11 years' time?

Answer: The delivery van will be worth R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.


STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

the value now=94 000rate of depreciation=6%=0,06number of years=11the final value=?

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

A=P(1in)A=94 000(1(0,06×11))

STEP: Solve for A
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=94 000(10,66)=94 000(0,34)=31 960

In 11 years' time the delivery van will be worth R31 960.


Submit your answer as:

Simple depreciation

A delivery van is worth R33 000 now. It depreciates at a rate of 4% p.a. following the straight-line depreciation method. What is the delivery van worth in 8 years' time?

Answer: The delivery van will be worth R .
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.


STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

the value now=33 000rate of depreciation=4%=0,04number of years=8the final value=?

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

A=P(1in)A=33 000(1(0,04×8))

STEP: Solve for A
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=33 000(10,32)=33 000(0,68)=22 440

In 8 years' time the delivery van will be worth R22 440.


Submit your answer as:

Compound depreciation

Farmer Azubuike buys a truck for R234 000. The truck depreciates by 12% per year on a reducing-balance basis. What is the depreciated value of the truck after 5 years?

Answer:

The depreciated value of the truck is R .

numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

original value=R234 000rate of depreciation=12%=0,12number of years=5the final value=?

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

A=P(1i)nA=234 000(10,12)5

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=234 000(0,88)5=234 000×0,52773...=123 489,2685...

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the truck after 5 years is R123 489,27.


Submit your answer as:

Compound depreciation

Farmer Chibuzo buys a plough for R139 000. The plough depreciates by 14% per year where the depreciation happens at a compound rate. What is the depreciated value of the plough after 7 years?

Answer:

The depreciated value of the plough is R .

numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

original value=R139 000rate of depreciation=14%=0,14number of years=7the final value=?

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

A=P(1i)nA=139 000(10,14)7

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=139 000(0,86)7=139 000×0,34792...=48 361,9672...

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the plough after 7 years is R48 361,97.


Submit your answer as:

Compound depreciation

Farmer Rethabile buys a plough for R169 000. The plough depreciates by 8% per year on a reducing-balance basis. What is the depreciated value of the plough after 8 years?

Answer:

The depreciated value of the plough is R .

numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!


STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

original value=R169 000rate of depreciation=8%=0,08number of years=8the final value=?

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

A=P(1i)nA=169 000(10,08)8

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

A=169 000(0,92)8=169 000×0,51321...=86 733,9895...

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the plough after 8 years is R86 733,99.


Submit your answer as:

Simple interest: finding i

Sehlolo deposits R3 000 into a savings account which gets simple interest. The account grows to a value of R5 727,00 in 9 years. Determine the value of i, the rate of interest paid by the bank.

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

  • A=5727,0
  • P=3 000
  • n=9
  • i=?

The question says, "gets simple interest," so we need to use the simple interest formula. Substitute in what you know!

A=P(1+in)5727,0=3 000(1+i(9))

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for i. The first step is to divide by 3 000 which will let us get into the brackets. Then solve it!

5727,0=3 000(1+9i)5727,03 000=(1+9i)1,909=1+9i0,909=9i0,9099=i0,101=i

Finally, multiply by 100 to change into a percentage. Then round to one decimal place if necessary. i=100(0,101)=10,1%.


Submit your answer as:

Simple interest: finding i

Mike deposits R6 000 into a savings account which pays simple interest. The account grows to a value of R8 982,00 in 7 years. Determine the value of i, the rate of interest paid by the bank.

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

  • A=8982
  • P=6 000
  • n=7
  • i=?

The question says, "pays simple interest," so we need to use the simple interest formula. Substitute in what you know!

A=P(1+in)8982=6 000(1+i(7))

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for i. The first step is to divide by 6 000 which will let us get into the brackets. Then solve it!

8982=6 000(1+7i)89826 000=(1+7i)1,497=1+7i0,497=7i0,4977=i0,071=i

Finally, multiply by 100 to change into a percentage. Then round to one decimal place if necessary. i=100(0,071)=7,1%.


Submit your answer as:

Simple interest: finding i

Kabir deposits R4 000 into a savings account which earns simple interest. The account grows to a value of R5 440,00 in 6 years. What is the interest rate for this account?

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.
Answer: % per annum.
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

  • A=5440,0
  • P=4 000
  • n=6
  • i=?

The question says, "earns simple interest," so we need to use the simple interest formula. Substitute in what you know!

A=P(1+in)5440,0=4 000(1+i(6))

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for i. The first step is to divide by 4 000 which will let us get into the brackets. Then solve it!

5440,0=4 000(1+6i)5440,04 000=(1+6i)1,36=1+6i0,36=6i0,366=i0,06=i

Finally, multiply by 100 to change into a percentage. Then round to one decimal place if necessary. i=100(0,06)=6,0%.


Submit your answer as:

3. Tax

Exercises

4. Annuities

Exercises

5. Amortisation

Exercises